Algebra
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Real algebra questions asked by students, solved step by step Every question below was submitted by a real student and answered step by step.
Map W₂ into the dual of W₁ by x ↦ ⟨x, ·⟩. Its image has dimension at most dim W₁, so rank-nullity forces the kernel W₂ ∩ W₁⊥ to have dimension at least dim W₂ − dim W₁ ≥ 1.
There is no solution at all. Simplifying the right side gives |x| = -5, and an absolute value is a distance from zero, so it can never equal a negative number.
The only solution is x = -5/2. When an absolute value equals zero there is no case split, because zero is the only number whose distance from zero is zero.
The solutions are x = 0 and x = 10/3. Because |A| = 5 forces A = 5 or A = -5, this single absolute value equation becomes two ordinary linear equations.
The two solutions are x = −2/3 and x = −2. An absolute value equal to a positive number produces two cases, 3x+4 = 2 and 3x+4 = −2, each solved in one step.
The solutions are x = -1/4 and x = -5/4. Split |4x + 3| = 2 into the two cases 4x + 3 = 2 and 4x + 3 = -2, then solve each linear equation separately.
The solutions are x = 0 and x = 10. Split the equation into 5 - x = 5 and 5 - x = -5; the negative coefficient of x is what makes the second case give x = 10.
The solutions are x = 3.4 and x = -3.4. The minus sign inside the bars changes nothing, because a number and its opposite are the same distance from zero.
The solutions are x = 5 and x = 3/23. Pull the constants outside the bars, cross-multiply to 8|2x − 3| = 7|x + 3|, then square to get 23x² − 118x + 15 = 0.
The solutions are x = -1 and x = 2. Splitting at x = 0 and x = 1 gives three cases; the middle case collapses to 1 = 3 and therefore has no solution.
There are four solutions: x = ±4 and x = ±√10. An absolute value wrapped around a quadratic splits into x²−13 = 3 and x²−13 = −3, each giving two roots.
The answer is -1 <= x <= 2. A less-than-or-equal absolute value becomes one closed double inequality, -3 <= 2x - 1 <= 3, solved by adding 1 and halving.
The answer is -9 < x < 4. A less-than absolute value splits into one double inequality -13 < 2x + 5 < 13, which you solve by subtracting 5 and halving.
The answer is w <= 40/7 or w >= 44/7. Divide by 7 first, then a greater-than-or-equal absolute value splits into two outward rays instead of one interval.
The solution is w <= 3 or w >= 9. Divide by 7 to reach |w - 6| >= 3, then read that as: w sits at least 3 units away from 6, in either direction.
The solution is a < -1 or a > 1/3. With absolute values on both sides, squaring is reversible, turning the problem into the quadratic 3a^2 + 2a - 1 > 0.
The solution is m < -9 or m > 9. Isolate |3m|, divide by -2 and flip the sign to get |3m| > 27, then split the outward inequality into two branches.
The solution is all real numbers. After dividing by -2 the inequality reads |3m| > -24, and since an absolute value is never negative it always holds.
The answer is x <= (11 - sqrt 97)/4 or x >= 5, about x <= 0.288 or x >= 5. Breakpoints at 2, 3 and 4 give four cases; only the outer two have solutions.
The solution is 4/3 <= x <= 2. A variable right-hand side must first be forced non-negative, and then the sandwich -(x-1) <= 2x-3 <= x-1 is solved.
The solution is x < -2, x = 1, or x > 2. The numerator is never negative, so the sign follows the denominator, and x = 1 makes the whole fraction exactly zero.
The solution of (|x − 2| − x)/(x² − 4) ≤ 0 is (−2, 1] ∪ (2, +∞). Split the absolute value at x = 2, build a sign table, and see why x = 1 is in but ±2 are out.
Answer: sqrt(2), about 1.414214. Adding the like terms gives 2/sqrt(2), and rationalizing by sqrt(2)/sqrt(2) cancels the 2 to leave sqrt(2) exactly.
The sum is 7 times the fifth root of 4. Because both terms already share the same index and radicand, only their coefficients 2 and 5 are added together.
The sum equals 37 times the square root of 2. Simplify root 50 into 5 root 2 and root 32 into 4 root 2, then add the resulting coefficients 25 and 12.
The result is 7x^2 - 7x + 9 with degree 2. Distribute the subtraction sign across the third polynomial first, then collect the x^2, x and constant terms separately.
The sum is -x^3 + 8x^2 - 15x - 4 and its degree is 3. Add the coefficients of matching powers; the cubic terms do not cancel, so the degree stays 3.
The sum is −4x² − 11x + 13. See how to line up like terms by degree, why the lone −11x carries straight through, and how to check the result at x = 1.
(f + g)(x) = 6x - 4 with domain all real numbers. Stack the two rules, add the x-terms 5x + x = 6x and the constants 2 - 6 = -4, then state the domain.
(f + g)(x) = 4x - 58, with domain all real numbers. The two x^2 terms are exact opposites and cancel, so the sum of two quadratics comes out linear.
The sum equals sqrt(10). Squaring gives 6 + 2sqrt(9-5) = 10, and since both radicals are positive the sum is the positive root, not -sqrt(10).
The answer is 6x^2 + yz. Dropping the brackets is safe because the groups are added, not subtracted, and the two z^2 terms cancel exactly against each other.
The sum is 2(x^2 - 8x + 32) / (x(x - 8)). Cross-multiplying onto the denominator x(x - 8) gives x^2 + (x - 8)^2, which expands and factors to 2(x^2 - 8x + 32).
The sum is (x^2+13x+10)/(2(x-1)(x+2)). Factor 2x-2 as 2(x-1) and x^2+x-2 as (x+2)(x-1) to see the least common denominator, then build both numerators up to it.
There were 108 adult and 48 child tickets, rising to 128 adult. Writing the counts as 9k and 4k keeps one unknown, and 3(9k + 20) = 8(4k) gives k = 12.
The older brother is 25 3/5 years old. Let the younger age be x, write 8x + 8 = 3(x + 8), solve 5x = 16, then multiply back to get the older age.
Answer: the older brother is 128/5 = 25.6 years old and the younger is 16/5 = 3.2. Setting the younger age to x gives 8x + 8 = 3(x + 8), so 5x = 16.
The child is 18 and the father is 72. The equation 4x - 12 = 10(x - 12) gives x = 18, and twelve years ago the ages were 6 and 60, a ratio of exactly 10.
The older sibling is 25 and the younger is 5. Setting 5a + 5 = 3(a + 5) gives 2a = 10; both siblings gain the same 5 years, which is what makes the ratio fall.
I am 20 and my mother is 60. Subtracting 10 from both ages gives 3x - 10 = 5(x - 10), so x = 20; ten years ago the ages were 10 and 50, a ratio of 5.
The ages are 88/3 and 50/3 years, about 29.33 and 16.67. Two conditions give two equations; substitution reduces them to 3D - 4 = 46.
The area is 140 m². Setting the width to x makes the length x + 4; solving 2(2x + 4) = 48 gives x = 10, so the sides are 10 m and 14 m.
The area is 160 m². Let the width be x, write the length as x + 6, solve 2(x + (x+6)) = 52 to get x = 10, then multiply 10 by 16.
From u_8 = u_1 + 7d we get 7d = 26 - 1/3 = 77/3, so d = 11/3. Substituting back gives u_8 = 1/3 + 7(11/3) = 78/3 = 26, confirming the fraction.
Since u_4 = 10 the second equation gives u_6 = 16. Because u_6 - u_4 = 2d, we get 2d = 6 and d = 3, with u_1 = 1 reconstructing the whole sequence.
From u_4 - u_2 = 2d we get d = 2 and u_1 = 1. Then u_15 = 1 + 14(2) = 29, so the fifteenth term is 29 and not the tempting 31.
The nth-term rule u_n = u_1 + (n - 1)d with u_1 = 3 and d = 2 gives u_5 = 3 + 4(2) = 11. Listing 3, 5, 7, 9, 11 confirms the fifth term is 11.
The monthly cost is 30 + 0.05t + 0.10x dollars for t talk minutes and x texts sent. Evaluating at five usage patterns gives 40, 38.80, 41.25, 43.50 and 42.50 dollars.
The solution is x <= -9/2, that is the interval (-infinity, -9/2]. Split the chain into two inequalities, watch every x^2 term cancel, and intersect the results.
log 1576 / log 2.44 = log base 2.44 of 1576 = 8.2541. The shared base cancels, and raising 2.44 to the 8.2541 power returns 1575.99, confirming the exponent.
Answer: only the last two are equal. Expression 1 simplifies to 0.015S - 0.955 and expressions 2 and 3 both to 0.0138462S - 0.789231; they agree only at S = 431/3.
There are 30 five-unit coins and 30 ten-unit coins. Writing the second count as 60 - x turns the two conditions into the single equation 5x + 10(60 - x) = 450.
There are 30 five-unit coins and 50 ten-unit coins. The single equation 5x + 10(80 - x) = 650 encodes both the count and the value condition at once.
The expression equals (2x + 2)² = 4(x + 1)². Spot the A² + 2A + 1 pattern with A = 2x + 1 instead of expanding, then factor the 2 out of the bracket.
The answer is x, not 1x or 0. Adding the coefficients gives 4 + 2 - 5 = 1, and a coefficient of 1 is written invisibly in front of the variable.
The answer is -9xy^2. All three terms carry the identical variable part xy^2, so only the coefficients combine: 6 - 3 - 12 works out to -9, and xy^2 rides along.
The answer is x. All three terms share the same variable, so only the coefficients add: 9 - 14 + 6 = 1, and a coefficient of 1 is never written out.
The simplified form is 2x² + 7x − 7. Group terms by degree: 3x² − x² = 2x², then 2x + 5x = 7x, and the constant −7 has nothing to pair with.
The simplified form is -x^5 - 15x^4 - 6x + 1. Only the two x^3 terms are like terms and they cancel, which drops the degree-3 term entirely.
Answer: x = -12/5 = -2.4. Group the three x-terms into -5x and the three constants into -12, then divide to get x = -12/5, written as a fraction.
-12xy + 9xy - 3xy = -6xy. All three terms carry the identical variable part xy, so only the coefficients are added: -12 + 9 - 3 = -6, and xy is left alone.
The answer is x ≈ 0.0011432627. All three terms are like terms, so the coefficients add to 87,000,451.12 before a single division finishes the problem.
If a < b then 3 − 2a > 3 − 2b, and vice versa. Subtracting 3 from both sides is harmless, but dividing by −2 reverses the inequality.
If a < b then 4a − 3 < 4b − 3. Multiplying by the positive 4 and subtracting 3 both preserve the inequality, so the direction never changes.
a^3 + b^3 is always at least a^2b + ab^2, with equality only when a = b, because the difference factors as (a + b)(a - b)^2, a product of nonnegative pieces.
Kelly is wrong: both initial values are -1. Function H has slope 1/2 through (8, 3), so tracing back to x = 0 lowers y by 4, landing on -1 just like G.
The answer is 3(x - 2/3)^2 + 2/3, minimum 2/3. The discriminant is -8, so the zeros are the complex pair (2 plus or minus i root 2)/3, not real numbers.
The answer is (x - 1)^2 + 2, so the minimum value is 2 at x = 1. The discriminant is -8, which is why this quadratic never factors over the real numbers.
Answer: no real solutions; x = -0.19717 +/- 0.13382i. Move -0.2 across, get 4.3957x^2 + 1.7334x + 0.2496 = 0, and read the negative discriminant -1.38399.
(g∘f)(x) = 128x² − 456x + 403, (h∘f)(2) = 11 and (f∘h)(x) = 16x³ + 36x − 15. The inner function is substituted wherever the outer function's variable appears.
The solution is -3 < x <= -2, or (-3, -2]. Subtracting 4 from all three parts of the compound inequality isolates x in a single move without flipping signs.
The solution is 1/6 < x < 17/60. Multiplying by 7, dividing by 60 and subtracting 1 — all positive operations — keeps both inequality signs pointing the same way.
The answer is 17/60 < x < 2/5. Multiply all three parts by 7, divide by 60, then subtract 1 - every step keeps the direction because both factors are positive.
The solution is 8 < y <= 16. Add 8 to all three parts, then divide all three by 3; because 3 is positive the inequality signs and their strict/non-strict status never change.
The solution is x > 7/5. The variable sits in both outer parts, so the chain must be split into two inequalities and their solution sets intersected.
The answer is -11/60 < x < -1/15. Both bounds come out negative because 60/7 is already about 8.57, so 1 + x has to be pulled below 1 to fit.
There is no solution. The inequality demands a number greater than 8 and less than 7 at once, and the algebra confirms it by producing reversed bounds.
The answer is -1/15 < x < 1/20. This is the window that straddles zero, because 60/7 is about 8.57 - already inside the target range at x = 0.
There is no solution. Once 4018 - 2734 is simplified to 1284, the chain demands 3818 ≤ 499x - 20y < 1284 and 1284 ≤ 33x - 78y < -2534, both impossible.
The answer is -1 <= x <= 3. Work on all three parts at once: add 1 everywhere to get -3 <= 3x <= 9, then divide by the positive 3 with no sign flip.
Answer: 15/2 < x <= 21/2, that is (7.5, 10.5]. Multiply all three parts by 12 to get 2 < 2x - 13 <= 8, add 13, then halve - the strict and closed ends are preserved.
A = x^5(x - 1)^5 / (x^2 + 1)^4. The power rule moves each coefficient into an exponent, and the negative coefficient sends its factor into the denominator.
Answer: 7 integers (-4, -3, -2, -1, 2, 3, 4). Factor both parts, note the denominator opens downward, and read the sign chart between the critical points -4, -1/5, 1 and 4.
There are 15 similarity classes. Over ℂ the Jordan form is a complete invariant, so the count is p(3)·p(4) = 3·5 = 15, one for each pair of integer partitions of the two multiplicities.
Answer: x = -59/3, about -19.667. Cube both sides to get 5 - 3x = 64; unlike a square root, a cube root needs no domain condition and creates no extraneous roots.
The real solution is x = −33. Cube both sides to clear the 1/3 exponent, solve 2x + 2 = −64, and see why an odd root never creates extraneous solutions.
The three real roots are x ≈ −7.4802e−8, 0.00106167 and 0.00814390. Scaling x by 10³ first turns the badly conditioned decimals into a well-behaved cubic.
Answer: x = 1 and x = (3 +/- sqrt(21))/2, about 3.7913 and -0.7913. The key step is spotting that 214221 = 101^2 x 21, so the square root simplifies exactly.
One real root s ≈ −3.086409 and a complex pair −6.706796 ± 1.902385i. No rational root exists, so depress to y³ − 3y/4 − 49/4 and apply Cardano.
Answer: x = 1.3210837, the only real root. The discriminant -42,120,000 is negative, so Cardano gives one real root and two complex ones; no rational root exists.
Answer: x = -24.8617, -2.4502 and 2.2304. Vieta is the fastest check: the three roots must sum to -92.3/3.68 = -25.0815 and multiply to 500/3.68 = 135.87.
The three real roots are x ≈ 0.4797, 8.4956 and 12.0248. No rational root exists, so depress the cubic to y³ − 35y + 49 and use the casus irreducibilis cosine formula.
The solution set is (-inf, -2) U (1, 3). Find the three zeros, split the number line, and read the sign of the product on each piece with a sign chart.
The solution set is (−∞, −1) ∪ (−1, 2). Factoring out x³ + 1 gives (x+1)²(x²−x+1)(x−2) < 0, and only x − 2 can be negative, with x = −1 excluded.
Answer: (-inf, -1) union (2, 4). Three simple roots split the line into four intervals whose signs alternate, and a strict inequality excludes all three roots.
The polynomial is f(x) = 5(x + 3)(x - 6)(x - 10) = 5x³ - 65x² + 60x + 900. See how each zero becomes a factor and how the leading coefficient fixes the scale.
The only real root is x ≈ 0.00548128. Recognising 0.016875 = 3(0.005625) and 9.4921875e−5 = 3(0.005625)² collapses the cubic to a shifted perfect cube.
Answer: x = -6. The cubic is a near-miss for (x+4)^3, whose constant is 64; the extra 8 makes it (x+4)^3 = -8, so x + 4 must equal the real cube root -2.
Roots: x = -1.228936, 1.348310 and 13.880626. The absent x term makes the pairwise product sum zero, which is the sharpest check on any numerical answer here.
Roots: x = -0.012650903, 0.023902272 and 245.699748632. Drop x^3 near zero to get the two small roots, then use the root sum 245.711 for the large one.
Roots: x = -0.014681475, 0.025933162 and 245.699748313. Drop x^3 near zero to get the two small roots, then use the root sum 245.711 for the large one.
Roots: x = -0.016747308, 0.027999354 and 245.699747954. Drop x^3 near zero to get the two small roots, then use the root sum 245.711 for the large one.
Roots: x = -0.018742545, 0.029994971 and 245.699747574. Drop x^3 near zero to get the two small roots, then use the root sum 245.711 for the large one.
The only real solution is x = 2. The cubic is (x − 1)³ − 1, so the equation becomes (x − 1)³ = 1 and the real cube root gives x − 1 = 1.
Roots: x = -15.219737, 11.273122 and 62.946615. No rational root divides 10800 into a zero, so bracket each sign change and refine with Newton's method.
The only real solution is x = 2 + ρ ≈ 3.3247180, where ρ is the plastic number. Factor the left side as (x−1)(x−2)(x−3), substitute y = x − 2, and solve y³ − y − 1 = 0.
The only real root is x ≈ −28.392868, with complex pair −5.803566 ± 6.062907i. Depressing gives y³ − 400y/3 + 38000/27 and a positive discriminant.
The cyclist takes 1.5 hours, riding at 20 km/h. Setting 30/v - 30/(v + 40) = 1 clears the fractions to v^2 + 40v - 1200 = 0, whose only positive root is 20.
The sum of the roots is 10. Subtract the linear pattern 4x - 3, factor the difference as a(x-1)(x-2)(x-3)(x-4), and apply Vieta - the constant a cancels.
The equation has no real solutions. The polynomial is exactly (x⁴ − x³ + x² − x + 1)² + 1, and a real square can never equal −1, so the value is never below 1 (its actual minimum is about 1.4537).
The answer is a = (root 5 - 1)c/4. The messy radical root(6 - 2root 5) denests to root 5 - 1, because 6 - 2root 5 is the perfect square of that difference.
The nested radical equals √6 − √5 ≈ 0.2134. Write 11 − √120 as 11 − 2√30 and find m + n = 11 with mn = 30, giving the perfect square (√6 − √5)².
The solution is the unbounded region with corner (2,4), bounded left by x = 2 and above by y = 6 - x. Solving each inequality for y makes the overlap easy to shade.
The determinant is -16. Expand along the top row: 9(32-40) - 2(112-144) + 2(140-144) = -72 + 64 - 8 = -16. Changing one row entry from 27,7,7 to 28,8,8 moves the value from -9 to -16.
The determinant is -9. Expand along the top row: 9(28-35) - 2(108-126) + 2(135-126) = -63 + 36 + 18 = -9.
The determinant is 9. Subtract multiples of the third row to clear the first column down to a single entry 9, then expand along that column and evaluate a 3x3 that comes to 1.
The determinant is -46. Two rounds of row operations reduce the first column to a single 1 in the top-left corner, leaving a 3x3 determinant of -46 to evaluate.
The answer is 8x. Instead of expanding both squares, use A² − B² = (A − B)(A + B) with A = 2x + 1 and B = 2x − 1 so the quadratic terms never appear.
4(x + 3)(x - 3) = 4(x^2 - 9) = 4x^2 - 36. The two brackets are a conjugate pair, so a^2 - b^2 collapses them in one step instead of a full FOIL expansion.
The expression collapses to -16a(a^2+2)(a^2+8)/(a^2-4)^2. Recognising a^3+8 and a^3-8 turns each square into a single fraction before subtracting.
The answer is (2x + 3m)(2x − m)/48. Each term is already a perfect square, so a² − b² applies directly with a = x/3 and b = (2x/3 − m)/4, i.e. x/6 − m/4.
The answer is 8xy. Factoring as a difference of squares gives the factors 4y and 2x, so all the squared terms cancel before you ever expand anything.
Each 20-term block is a perfect cube, so the difference is A^3 - B^3 with A - B = (c - e)(y - b). Recognising the cube collapses 40 terms into three factors.
Answer: 0. Each fraction reduces to -(3a + 2), because the reversed denominators supply a factor of -1, so the difference cancels identically.
The difference quotient is 10x + 5h − 3. Expanding f(x+h) cancels every term without an h, so the h in the denominator divides out cleanly.
The result is −21/((3x+3h−1)(3x−1)). Putting the two fractions over a common denominator makes the numerator collapse to −21h, so the h cancels.
The difference quotient simplifies to 2x + h - 5. See how to expand f(x + h), subtract f(x), factor h out of the numerator and cancel it, step by step.
The quotient simplifies to −2x − h + 7, so A = −2, B = −1 and C = 7. All the terms without an h cancel, which is why the h in the denominator can be divided out.
The result is 5/(√(5x+5h+1) + √(5x+1)). Multiplying by the conjugate turns the difference of roots into 5h, letting the h in the denominator cancel.
OP = 1. Substituting t = m^2 + n^2 turns the condition into t^2 + 4t - 5 = 0, whose roots are 1 and -5; only t = 1 survives because a sum of squares cannot be negative.
Answer: 2m - 7 + 31/(m + 5). Reorder into 2m^2 + 3m - 4, synthetic-divide with the root -5, and check that the numerator at m = -5 really equals the remainder 31.
Answer: x^2 - 2 with remainder 0. Long division clears in two steps, and grouping confirms it, since 2x^3+x^2-4x-2 equals x^2(2x+1) - 2(2x+1) exactly.
The quotient is x - 1/2 and the remainder is -1/2. Insert the missing 0x term, then divide leading terms twice; the non-monic 2x forces fractional quotient terms.
The quotient is x - 1/2 and the remainder is 5/2. Two rounds of dividing leading terms finish it, and the division identity confirms the fractional pieces.
Answer: 2x + 1, for x not equal to -3. Split the middle term using 6 and 1, factor by grouping into (2x+1)(x+3), then cancel the common binomial factor.
Answer: 3p + 7 + 5/(p - 1). Reorder into descending powers first, then synthetic-divide by the root 1; the remainder 5 equals the value of the numerator at p = 1.
The quotient is 3x - 3 and the remainder is -2. Insert the missing 0x term before dividing, then confirm with the remainder theorem by evaluating at x = -1.
The quotient is 3x³ + 6x² + 8x + 24 with remainder 47. See why the missing x³ term has to be written as 0x³ before dividing, and how to check with f(2) = 47.
Factor 4a^2 - b^2 as (2a-b)(2a+b); the (2a+b) cancels with (b+2a), leaving (2a-b)/(b-2a)^2, which simplifies to 1/(2a-b).
Answer: 4p + 3 + 10/(p - 2). Reorder to 4p^2 - 5p + 4, divide synthetically by the root 2, and confirm the remainder by evaluating the numerator at p = 2.
Factor 9 - a^2 as -(a-3)(a+3) to expose the sign, then cancel (a-3) and (a-5) after inverting the divisor. The minus sign survives into the final answer.
The quotient is 2x^3 + x^2 - 4x - 2 with remainder 0, so x = 1 is a root. The bottom row of synthetic division gives the coefficients directly.
All four quadratics factor: (a-6)(a+3), (2a+1)(a-6), (a+3)(a-2) and (2a+1)(a-1). After inverting, three pairs cancel and only (a-1)/(a-2) remains.
The quotient is 5xy + 4x - 3y. Split the fraction into three, divide coefficients and subtract exponents, then multiply back to confirm the original numerator.
Answer: f^2 - 4f + 16, with no remainder. Write the missing f^2 and f terms as zeros, run synthetic division with -4, and recognise the sum-of-cubes factorisation.
Answer: j^2 + 4j + 16, remainder 0. Use coefficients 1, 0, 0, -64 with the root +4; this is the difference-of-cubes identity, so every sign in the quotient is positive.
The result is x^2 - x + 1 - 1/(x + 1). Long division on x^3 + 0x^2 + 0x + 0 gives quotient x^2 - x + 1 and remainder -1, which matches the remainder theorem.
The quotient is x² − 2x + 3 with remainder 0, so x − 1 is a factor. Follow all three divide-multiply-subtract rounds and the factor-theorem check.
The quotient is x + 2 with remainder 0. Learn why dividing by a quadratic gives a degree-1 quotient and only takes two rounds, with a factoring cross-check.
Answer: x + 5, with remainder 0. Long division cancels exactly, or factor by grouping: x^4+5x^3-3x-15 = (x+5)(x^3-3).
The quotient is x/3 - 5 and the remainder is 50. Dividing x^2 by 3x gives the fractional lead term x/3, after which the second step clears completely.
The quotient is x³ + 2x² + 17x + 20 with remainder −80x − 148. Reorder both polynomials by degree, insert zero placeholders, then run four divide-multiply-subtract rounds.
The quotient is (y - 9)(y^2 + 9)/(y - 5). Multiply by the reciprocal, factor the difference of squares and the trinomial, then cancel the shared (y + 9).
Answer: A = 5. Dividing 2x^3 - x^2 + 2x + 5 by x + 1 gives the quotient 2x^2 - 3x + 5, so the third (constant) quotient entry labelled A is 5.
The domain is the open interval from -2 to infinity. The radicand must be nonnegative and the denominator nonzero, and together those force x + 2 > 0 strictly.
The domain is all reals except 5 and 7, written as three intervals. Factoring the denominator as (x - 5)(x - 7) shows exactly which two values make it zero.
The domain excludes x = ±1 and ±2, giving five open intervals. The quartic denominator factors via u = x² into (x²−1)(x²−4).
The domain is (−∞, −12) ∪ (−12, 15) ∪ (15, ∞). Each denominator contributes one excluded value, splitting the real line into three open intervals.
The domain of f(x) = 1/(8/(x − 5) − 2) is (−∞, 5) ∪ (5, 9) ∪ (9, ∞). A complex fraction hides two restrictions: the inner denominator and the outer one.
The domain excludes -3, -1 and 1. Factoring by grouping turns the cubic into (x + 3)(x - 1)(x + 1), so the real line breaks apart into four open intervals.
The domain is (-infinity, infinity). Both f(x) = 2 - x^2 and g(x) = x^2 + 4x - 60 are polynomials defined for every real number, so their difference is too.
The domain is [4, infinity). Each radicand must be nonnegative, giving x at least 4 and x at least -15, and the intersection keeps only the stricter condition.
The domain is [-10, infinity). A square root needs a nonnegative radicand, so x + 10 >= 0 gives x >= -10, and -10 itself is included since sqrt(0) = 0.
The domain is (-inf, inf). A polynomial has no denominator, no even root and no logarithm, so every real input is legal; the discriminant only limits the range.
The domain is [3, ∞). A square root needs a nonnegative radicand, so 4x − 12 ≥ 0 gives x ≥ 3 — and the endpoint is included because √0 = 0 is defined.
The domain is (−∞,−6) ∪ (−6,6) ∪ (6,∞). A cube root is defined for every real number, so only the zeros of the radicand — where the denominator vanishes — are excluded.
The solution is (−1, 0) ∪ (1, 2). Split into two quadratic inequalities, solve each with a sign chart, then intersect — the strict quadratic side punches a hole at [0, 1].
a = b is the correct conclusion. Combining b^2 = 4ac with 2b = a + 4c yields (a - 4c)^2 = 0, so a = 4c, and substituting back gives b = 4c as well.
The only root is x = −0.1. Equal squares force 5x + 7 = ±(5x − 6); the plus case gives the impossible 7 = −6, and the minus case gives 10x = −1.
The line is y = x/3 - 25/3. Two negatives in the slope give the positive 1/3, and the fractional intercept should be kept exact rather than rounded.
The line is y = 2x + 13. Subtracting a negative x-coordinate gives the run 3, and the steep slope 2 pushes the intercept well above both given points.
Every real number is a solution. Because 1 - x and x - 1 are opposites, their squares are equal, so the two sides are literally the same expression.
The solution is x = 11. Expanding both sides makes the x^2 terms equal (both 1/2 x^2), so they cancel and a linear equation in x is left behind.
The solution is x = 44. The x^2/12 term appears on both sides after expanding (x+3)^2, so it cancels and only a linear equation in twelfths and twentieths is left.
The value is 514. Simplify inside the bracket to get 4, cube it to get 64, multiply by 8 to reach 512, and only then add the 2 sitting outside.
The value is 13/2. Adding the two cube terms first gives 20x^3 + 6x + 1, so only one power of 1/2 has to be computed.
The value is 18. Difference of squares turns the first two terms into (4x + 2)(4) = 16x + 8, so the whole expression is 10x + 8, which is 18 at x = 1.
The value is 62. Expanding first collapses the expression to 20x + 61, so the awkward fraction x = 1/20 never has to be squared.
Combining 8x³ and 12x³ gives 20x³ + 6x + 1, which at x = 1/2 equals 5/2 + 3 + 1 = 13/2. The near-miss 8x³ + 12x² + 6x + 1 = (2x + 1)³ would give 8 instead.
The value is 400. Both 9x² and 4y² are squares, and 12xy is exactly twice their roots multiplied, so the expression collapses to (3x − 2y)² = 20².
Set x = 2k and y = 3k. Every term is degree 2, so k^2 cancels: the numerator becomes 45k^2 and the denominator 29k^2, giving 45/29.
The value is about -65.63. Keep ln(90.1) = 4.500920 to six places: slipping to a coarse 4.5017 shifts the answer to -65.65, an error of about 0.02.
Answer: f(4) = 9. First check which condition x = 4 satisfies: it lies in 3 <= x <= 10, so the middle rule x^2 - 7 applies, giving 16 - 7 = 9.
g(-1) = -6, g(7) = 38 and g(0) = -4, while g(-2) is undefined. The first branch requires -2 < x, a strict inequality that excludes the endpoint -2 itself.
The value is 0. Substituting m = 3 and n = 2 gives 3 - 18/6 = 3 - 3, and cancelling the n first shows the expression is really m - m^2/3, which is zero at m = 3.
Answer: 50. Expand the sigma notation into 3 + 8 + 15 + 24, or use the closed forms for the sum of squares minus the count of terms - both give 50.
The value is 44. Substitute first, clear the parentheses and the exponent before multiplying, and only then subtract: 64 - 5(4) = 64 - 20 = 44.
The value is 10. Substituting x = 7 gives 49 - 42 + 3, so square before multiplying, keep 6x as one subtraction, and work left to right at the end.
The value is 25. The expression is the perfect square (x − 4y)², so the given relation can be substituted whole — no need to find x and y individually.
The answer is y = 4. Substitution is the whole method: replace x by its value in the second equation, giving 2 times 2, which is 4 rather than 2 plus 2.
The value is 29 + (142 - root 3)i. Cubing 3 - 2i gives -9 - 46i, and the -3 multiplier flips it to 27 + 138i, which dominates the real and imaginary totals.
Use the rank normal form PAQ = D with D the r-block identity. Taking B = QD⁻P, where D⁻ is the n×m block identity, gives ABA = A for every matrix A.
The answer is 7x^2 - 21x + 21. Expand the product with FOIL and the square with the (a - b)^2 rule, then add the two results together term by term.
The result is x^6 + 2x^5 + 3x^4 + 3x^3 + 3x^2 + 2x + 1. Squaring produces the palindromic pattern 1, 2, 3, 4, 3, 2, 1 and subtracting x^3 drops the middle 4 to 3.
The product is 40x⁶ − 68x⁴ − 58x³ − 24x² − 91x − 72. Distribute each of the three left terms over the trinomial, then collect the nine results by degree.
The product simplifies to 4x² − 6x + 1. Grouping (2x−1) as one block and √(2x) as the other turns the product into (2x−1)² − 2x.
(4x² − 3)² expands to 16x⁴ − 24x² + 9. Use the pattern (a − b)² = a² − 2ab + b² with a = 4x² and b = 3 instead of multiplying the terms out one pair at a time.
The expansion is 5x^3 + 5x^2 - 55x - 15. Multiply the two brackets first to get x^3 + x^2 - 11x - 3, then distribute the 5 - a step the usual distractor answers forget.
(515 - 0.27x) * 98^2 = 4946060 - 2593.08x. Square 98 to get 9604 first, then distribute: 515 * 9604 = 4946060 and 0.27 * 9604 = 2593.08 for the x term.
The product is 45x^4 + 54x^3 - 40x^2 - 48x. Both left-hand terms multiply both right-hand terms, and since all four degrees differ nothing combines.
The expansion is 5x^2y^3 - 10xy^2 - 25y^6 - 10y^5 + 25y^4 + 30y^3 - 25y^2. Only the y-exponents change on the terms without x, each rising by 2.
The product is 7x³ + 35x² − 21x − 49. Distribute 7x and then 7 across the trinomial, then combine the two like x² terms and the two like x terms.
The product is 35x^4 + 7x^3 - 39x^2 + 34x - 10. Distribute each of the three left-hand terms across the trinomial, then collect like powers of x.
(98 - 0.18x) * 270^2 = 7144200 - 13122x. Squaring 270 gives 72900, then 98 * 72900 = 7144200 and 0.18 * 72900 = 13122, a whole-number x coefficient.
The expansion is -48x^4y - 8x^3y^3 - 48x^2y^2 + 32x^2y - 40xy^4 - 56xy^2 + 64xy. Multiply 8xy into each of the seven terms, adding exponents on x and on y.
The numerator expands to a1x^2 - 2a1xy + (a1+a2)y^2 + b1x + (b2-b1)y. The nested form is Horner style, so unfolding it one bracket at a time is the safe route.
The cubic is -0.013122x^3 + 50.058x^2 - 57477.78x + 18572960. Multiply two binomials first, then distribute the third, tracking each decimal coefficient.
The result is -2 - 3 root 2. The product root 2 times root 2 collapses to 2, and the two remaining surd terms combine into a single -3 root 2.
The result is 2 root 6 - root 42. Distribute the radical, then multiply the two roots into one: root 6 times root 7 is root 42, which has no square factor.
The answer is m^2 - 4m + 4. The middle term is twice the product of m and 2, and it is negative; the last term is positive because a square is never negative.
The standard-form polynomial is 20x^3 - 6x - 42. Distribute each product, watch the double negative in -6 times -3x^3, then collect the two cubic terms.
The standard form is x^2/3 - 4x/3 - 5/3. Square the bracket first, distribute the 1/3, then combine 4/3 with -3 by writing 3 as 9/3 to get -5/3.
The answer is x^2/3 - 4x/3 - 5/3. Multiply the brackets first to get x^2 - 4x - 5, then distribute the 1/3 across all three terms, not just the first.
The answer is pq - p - q + 2. FOIL gives pq - p - q + 1, and the extra 1 outside the bracket lifts the constant to 2 rather than cancelling it.
Use (a+b+c)^2 = a^2+b^2+c^2+2(ab+bc+ca): the squares give 2+3+5 = 10 and the three cross terms give 2 sqrt6 + 2 sqrt10 + 2 sqrt15.
The answer is x^4 - 8x^(5/2) + 24x - 32/root x + 16/x^2. The fractional exponent makes each term drop by 3/2 rather than by 1, which is the easy slip.
The result is −8x³ − 48x² − 104x − 352, or −8(x³ + 6x² + 13x + 44). The quartic terms cancel because both binomials are raised to the same even power.
The product is x² − x − 6. The outer and inner terms −3x and +2x combine to −x, and the constants multiply to −6 because the signs differ.
The product is x^2 + 7x + 10. FOIL gives x^2, 5x, 2x and 10, and the two middle terms add to 7x because 2 and 5 sum to 7 and multiply to 10.
The product is x^4 + 4x^3 - 12x^2 - 32x + 64. Both factors are perfect squares, so the product collapses to (x^2 + 2x - 8)^2 and needs only one squaring.
Answer: x = 0 and x = 1. Write 9 as 3^2 and move the fraction upstairs, then the substitution t = 3^(x-1) turns it into the quadratic 3t^2 - 4t + 1 = 0.
The solution is -2 < x <= 6. Substitute t = 2^x, combine both sides over a single denominator, read the sign chart, then convert the bounds on t back into x.
The solution is ((3-sqrt17)/2, 0) union ((3+sqrt17)/2, infinity). Match the bases, flip the inequality because 2/5 < 1, then sign-chart (x^2-3x-2)/x.
The solution is 2 <= x < 4. Factor the numerator by grouping into (2^x - 2)(5^x - 25), factor the quadratic denominator, then read a four-region sign chart.
The solution is 0 < x <= log7(2) or log2(7) <= x < 4. Group 14^x = 2^x*7^x into (7^x - 2)(2^x - 7), factor -x^2 + 4x, and read the sign chart.
The solution is x = 0 or 1 < x < log2(3). Substituting t = 2^x collapses the three fractions into 2(t-1)^2/((t-2)(t-3)), a perfect square over a quadratic.
Expanding gives 12x + 12 = 7y - 7, so x = (7y - 19)/12. Substituting x = 2 gives 36 = 7y - 7 and y = 43/7.
The answer is (1/16)(x^3 + 8)^2. Substituting u = x^3 turns it into a quadratic that is a perfect square, and x^3 + 8 splits further as (x+2)(x^2-2x+4).
The factorization is (2x + 1)(5x − 4). Since 10 × (−4) = −40 and 5 + (−8) = −3, split −3x into 5x − 8x and group to pull out the shared binomial 2x + 1.
The factorization is (5x - 4)(25x^2 + 20x + 16). Write the two terms as (5x)^3 and 4^3, then apply the identity a^3 - b^3 = (a - b)(a^2 + ab + b^2).
The answer is 6(2x + 3). The GCF of 12 and 18 is 6, not 2 or 3, and after dividing both terms by 6 the bracket 2x + 3 has no common factor left.
The GCF is 4, leaving 3x^2 - 44x + 120, whose discriminant 496 is not a perfect square - so the roots are (22 plus or minus 2 root 31)/3, not 6 and 20/3.
18abc^2 - 12a^2b^2c = 6abc(3c - 2ab). The greatest common factor pairs gcd(18, 12) = 6 with the lowest power of each shared variable, namely a, b and c.
Answer: (3x - 7)(7x^2 - 6). The rational root search misses it because the only rational root is 7/3; grouping the terms in pairs finds the factorisation instantly.
The answer is (x - 2)(2x + 1)(x + 1). Grouping fails here, so the Rational Root Theorem finds x = 2, and synthetic division leaves 2x^2 + 3x + 1.
The answer is x(2 - 9y). Only x is shared by both terms, because 2 and 9 have no common factor and the variable y appears in the second term alone.
The factorisation is 2(x - 6)(x + 1). Every coefficient is even, so pulling out 2 leaves the monic x^2 - 5x - 6, which factors with the pair -6 and 1.
The factorisation is (x - 2)(2x + 1). Multiply 2 by -2 to get -4, split -3x as -4x + x, then group the four terms and pull out the shared binomial.
The factorisation is (2x + 5)(x - 4). The AC product is -40, and the pair 5 and -8 sums to -3, which splits the middle term ready for grouping.
Answer: not over the integers, but yes over the reals: 2x^2 - 4x + 1 = (root2(x - 1) - 1)(root2(x - 1) + 1). The discriminant 8 is positive but not a perfect square.
The factorization is (2x - 1)(x - 3y). Grouping the first two terms as x(2x - 1) and the last two as -3y(2x - 1) exposes a shared bracket to pull out.
The answer is 2x(x + 2). The greatest common factor combines the numeric part 2 and the variable part x, so both come out together, leaving x + 2 behind.
The factorisation is 8x(4xy^2 - 3). The numerical GCF of 32 and 24 is 8, and only one power of x is shared, since the second term has no y at all.
The factorisation is (x + 1)(x + 2)(3x - 2). Testing x = -1 gives zero, and the cubic quotient 3x^2 + 4x - 4 then splits by the AC method with 6 and -2.
The factorisation is (3x-5y+1)(x-3y+4). Factor the quadratic part into (3x-5y)(x-3y), then fit the two constants so the x and y coefficients come out as 13 and -23.
It factors as (3x^2 + 12x + 7)(x^2 + 4x + 9), giving real roots (-6 +/- sqrt 15)/3 and the complex pair -2 +/- i sqrt 5. Always expand a guessed factorisation back.
The answer is -B(2A - B)^2. After pulling out B the bracket is the negative of a perfect square, so a minus sign has to come out along with it.
The factorisation is 4(m - (1 + sqrt 6)/2)(m - (1 - sqrt 6)/2). The AC product -20 has no pair summing to -4, so the roots come from the quadratic formula.
The answer is 2mn(2m - 4n - 1). The last term is the whole GCF itself, so it leaves a -1 behind inside the bracket rather than disappearing.
The answer is (2x - 3)(2x + 3). Both terms are perfect squares, since 4x^2 is (2x)^2 and 9 is 3^2, so the roots of 4x^2 - 9 = 0 are x = 3/2 and -3/2.
It factors as (2x+y−3)(2x+y+1). The first three terms form (2x+y)², and the next two are −2(2x+y), so the whole thing is a quadratic in u = 2x+y.
The answer is y(2x + 3)^2. Pull out the common y first, and what remains is a perfect square trinomial because 12x equals 2 times 2x times 3.
5x⁴ − 80 factors as 5(x − 2)(x + 2)(x² + 4). Pull out the 5, apply the difference of squares twice, and stop once the sum of squares x² + 4 is left irreducible.
The factorisation is (3x-2a+3)(2x-a-4). Factor 6x^2-7ax+2a^2 into (3x-2a)(2x-a), then choose constants p and q with pq = -12 that also give the -6x and +5a terms.
The factored form is 3(x − y)²(x + y). Collapse the second term to 3(y − x)³, use (y − x)³ = −(x − y)³ to align both terms, then take out 3(x − y)².
The factorization is 3(5a³ − 7b⁸)(5a³ + 7b⁸). Pull out the GCF 3 first, then 25a⁶ and 49b¹⁶ become perfect squares (5a³)² and (7b⁸)², so difference of squares applies.
7x² − 7x − 140 factors as 7(x − 5)(x + 4). Remove the common factor of 7 first, then split the remaining trinomial x² − x − 20 into two binomial factors.
81x² − 121y² factors as (9x − 11y)(9x + 11y). Rewrite the two terms as (9x)² and (11y)², then apply the identity a² − b² = (a − b)(a + b) directly to both.
The complete factorization is 9x(x - 1)(x + 1). The greatest common factor is 9x, not just 9, and the leftover x^2 - 1 is a difference of squares.
The factorisation is (3x + 1)^2. Both outer terms are squares, and the middle term 6x is exactly twice the product of 3x and 1, which confirms the pattern.
The expression collapses to (a²+ab+b²)(x²+xy+y²). Expanding A², AB and B² and collecting x², xy and y² terms gives the same factor a²+ab+b² three times.
The cyclic sum factors as −(a − b)(b − c)(c − a)[3(a² + b² + c²) + 5(ab + bc + ca)], a much sharper result than the fully expanded degree-5 form.
The answer is (a-b)(a+x)^m(b+x)^(n-1). Pull out the lowest power of each base, subtract inside the bracket, and the x terms cancel to leave a-b.
The answer is (a^2 - 2ab + 2b^2)(a^2 + 2ab + 2b^2). Adding and subtracting 4a^2b^2 creates a perfect square, turning a sum into a difference of squares.
It equals (a^5 - 1)/(a - 1) and is irreducible over the rationals - but not over the reals, where it splits into two quadratics with golden-ratio coefficients.
The factored form is (1/12)x^2y^2(6x^2 + 4xy + 3y^2). The fractional GCF is 1 over the LCM of 2, 3 and 4, which clears every denominator at once.
The answer is (a^2+ab-1)(ab+b^2-1). Expand first, then notice ab+b^2-1 divides it, giving the symmetric pair (a(a+b)-1)(b(a+b)-1) shown here.
Answer: (x^2 + 6x - 6)(x^2 - 6). Pair the brackets so both products share the constant -6, then substitute y = x^2 + x - 6 to get a simple quadratic in y.
The factorization is (y′ − 2x² − x)(y′ − 2x² + x). Grouping the first three terms into (y′ − 2x²)² turns the expression into a difference of two squares.
It factors as 2(4x^2 - 3x + 3)(x^2 + 3), so the roots are (3 +/- i sqrt 39)/8 and +/- i sqrt 3. Both quadratics have negative discriminants, so there is no real solution.
Answer: (x+2)(3x^2-5)(x^2+x+1). Find the root x = -2, divide it out synthetically, then split the quartic 3x^4+3x^3-2x^2-5x-5 into two quadratic factors.
The answer is (a^2 + b^2)(a^4 - a^2b^2 + b^4). Rewrite a^6 as (a^2)^3 so the sum of cubes formula applies, even though the exponent 6 is even.
The factorization is (a - 2x + 3)(a + 2x - 3). Grouping the three x-terms reveals a perfect square, turning the whole expression into a difference of squares.
The factorisation is (a + 2)^2. The constant 4 is 2 squared and the middle term 4a equals 2 times a times 2, so the square-of-a-sum pattern applies exactly.
A sum of squares does not factor over the reals, but over C it becomes (a + bi)(a - bi), because the cross terms cancel and i squared equals minus one.
Answer: (x-1)^2 (x^2-3x+1). Expanding gives the palindromic quartic x^4-5x^3+8x^2-5x+1, which splits as (x^2+ax+1)(x^2+bx+1) with a = -2 and b = -3.
The factorisation is (x - 1)(x - 2) and the roots are x = 1 and x = 2. A positive constant with a negative middle term forces both factors to be negative.
The roots are x = 1 and x = 3. Find two numbers multiplying to +3 and adding to −4, giving (x − 1)(x − 3), then apply the zero-product property to each factor.
The factorization is (x + y)(4x²y² + a² + 12). Pairing the six terms three ways exposes the binomial x + y hiding inside each pair, which is then collected out front.
The factorization is (f - 4g + 3h)(f + 4g - 3h). The last three terms are exactly -(4g - 3h)^2, so the expression is the difference of squares f^2 - (4g - 3h)^2.
The complete factorization is -x^2 y(x - 1)(x - 5). Take out the GCF x^2 y first, then pull the leading minus out of -x^2 + 6x - 5 before splitting it.
The complete factorisation is 3x^6(2x^3 - 1)(2x^3 + 1). Taking out 3x^6 leaves 4x^6 - 1, which is still a difference of squares and must be split again.
The answer is (3c/4 - 1)(3c/4 + 1). The fraction 9/16 is a perfect square because both 9 and 16 are, so its square root is the fraction 3/4.
The factorisation is (1/7 - c^5 n^3)(1/7 + c^5 n^3). Both terms are squares, since 1/49 is (1/7)^2 and c^10 n^6 is (c^5 n^3)^2 by halving the exponents.
27x^3+y^3=(3x+y)(9x^2-3xy+y^2), x^3-8y^3=(x-2y)(x^2+2xy+4y^2), x^3+512=(x+8)(x^2-8x+64). Learn to spot the two perfect cubes and write the factors.
The answer is (x - 1)(x - 2)(x + 3). Test the divisors of 6 to find the root x = 1, divide out x - 1, then factor the leftover quadratic x^2 + x - 6.
The factorisation is (x - 1)(x + 2)^2. Testing x = 1 gives zero, synthetic division leaves x^2 + 4x + 4, and that quotient is a perfect square.
The answer is (x + 3)(x - 2)(x + 2). Grouping in pairs exposes the common factor x + 3, and the remaining x^2 - 4 is a difference of squares.
The answer is (x + 1)(x^2 + 2x + 5). Only x = -1 works among the rational candidates, and the quadratic factor has discriminant -16, so it stops there.
Over the rationals it is (x - 1)(x^2 + 5x + 5). The quadratic factor has discriminant 5, so only over the reals does it split further using (-5 +- sqrt 5)/2.
The answer is (x + 1)(x - 2)(x + 2). Grouping in pairs exposes the shared factor x + 1, and the leftover x^2 - 4 is a difference of squares.
The complete factorisation is (x + 1)^2 (x - 1). Grouping gives (x + 1)(x^2 - 1), and the second factor is itself a difference of squares that must be split.
It factors as (x + 1)(x^2 + x + 1)(x^2 - x + 1). The sum is the geometric series (x^6 - 1)/(x - 1), so factoring x^6 - 1 twice and cancelling x - 1 ends it.
The answer is (x + y)(x^4 - x^3y + x^2y^2 - xy^3 + y^4). A sum of powers factors only when the exponent is odd, and the long factor's signs alternate.
The quartic is (x² + x + 1)². Its palindromic coefficients 1, 2, 3, 2, 1 are the square of 1, 1, 1, and it has no real roots since x²+x+1 > 0 everywhere.
The factorization is (x² + xy + y²)². The palindromic coefficients 1, 2, 3, 2, 1 are the signature of squaring a three-term expression rather than a binomial.
The answer is (x^2 + 2x + 3)(x^2 - x + 1). The quartic has no rational roots, so match coefficients against (x^2+ax+b)(x^2+cx+d) with bd = 3 instead.
It factors as (x^2 - x + 1 - a)(x^2 + x + 1 + a). Completing (x^2 + 1)^2 rewrites the expression as (x^2 + 1)^2 - (x + a)^2, a plain difference of squares.
It equals 2(x - 2)(x + 2)(x^2 + 10). Adding the two fourth powers cancels every odd-degree term, leaving 2x^4 + 12x^2 - 80, which is a quadratic in x^2.
The factorization is 2x(x+4)(x^2+4x+14). Centering the substitution at y = x + 2 kills the odd powers, leaving a biquadratic in y that factors as a quadratic in y^2.
The simplified form is (x + 5)/(x + 6)^(6/5). With negative exponents the smaller one is -6/5, so that is the power you factor out of both terms.
The simplified form is -(x + 6)(x + 7)^(1/3). Factor out the lower power (x + 7)^(1/3), then simplify the leftover bracket 1 - (x + 7) into -x - 6.
The answer is (x - 64)(x - 36), and not (x - 48)^2. Evaluate 48^2 = 2304 first, then find the negative pair -64 and -36 that adds to exactly -100.
The factorization is (x - a - 1)(x - a + 2). Substituting u = x - a collapses the expression to u^2 + u - 2, which splits at once as (u - 1)(u + 2).
The answer is (x - y - 3)(x - y + 3). The first three terms collapse to (x - y)^2, and what remains is then a difference of squares against 3^2.
The answer is x(x - 4). Both terms contain x, so x is the greatest common factor; pulling it out leaves x - 4, and the roots are then x = 0 and x = 4.
The factorisation is (x-y-5)(x+y+1). Complete the square on the x terms and on the y terms so the expression becomes (x-2)^2 - (y+3)^2, then apply A^2 - B^2 = (A-B)(A+B).
The answer is (x - 3)(x - 4). A positive constant with a negative middle term means both numbers are negative: -3 and -4 multiply to 12 and add to -7.
The discriminant is -3, so there is no real factorisation. Over C it splits as (x - (1 + i root 3)/2)(x - (1 - i root 3)/2), a conjugate pair of roots.
x² + 25 is prime over the real numbers. A sum of squares has no real roots, so unlike x² − 25 it cannot be split into a product of two real linear factors.
The factorisation is (x+a+6)(x+a-1). Factor the constant term a^2+5a-6 into (a+6)(a-1) and notice their sum is exactly the coefficient 2a+5 of x.
The factorization is (x - 4y)(x + y + 2). Factor the degree-two part as (x - 4y)(x + y) first; the leftover 2x - 8y = 2(x - 4y) then shares that same bracket.
The factorisation is (x+2y-3)(x+3y+1). Factor the degree-2 part x^2+5xy+6y^2 into (x+2y)(x+3y), then find the two constants that reproduce the -2x and -7y terms.
The answer is (x + 4)(x - 3). A negative constant forces opposite signs, and 4 with -3 is the only pair that multiplies to -12 and adds to +1.
The answer is (x + 3)(x - 2). The negative constant forces opposite signs, and 3 with -2 is the only pair multiplying to -6 while adding to +1.
The answer is (x + a)(x - a + 1). The two numbers a and 1 - a multiply to a - a^2 and add to 1, which is exactly what the factoring pattern needs.
The discriminant is -3, so there is no real factorisation. Over C it is (x + 1/2 - i root 3/2)(x + 1/2 + i root 3/2), whose roots are cube roots of unity.
Answer: (x^2 - xy + y^2)^2. Expanding gives the symmetric quartic x^4-2x^3y+3x^2y^2-2xy^3+y^4, which factors as a square once you try (x^2+axy+y^2)(x^2+bxy+y^2).
The answer is (x - y)(x - 1)(x + 1). Since y - x = -(x - y), the two brackets match up, and the leftover x^2 - 1 is a difference of squares.
The factorisation is y(x - 2)^2. Pulling out the common y leaves x^2 - 4x + 4, a perfect square, so the answer needs both a GCF step and an identity step.
The answer is (x - 1)(x^104 + x^103 + ... + x + 1). A difference of nth powers always has x - 1 as a factor, whatever n is, and the cofactor is all plus signs.
The answer is (x + 1)(x^104 - x^103 + ... - x + 1). Because 105 is odd, x = -1 is a root, so x + 1 divides the expression and the cofactor alternates in sign.
The factorisation is (x^2 - 1/2)^2. Treating x^2 as the variable turns the quartic into a perfect square, since 1/4 is (1/2)^2 and -2 times x^2 times 1/2 is -x^2.
The answer is (x - y)(x^n-1 + x^n-2 y + ... + y^n-1). The cofactor's signs are all positive, and the middle terms telescope away when the product is expanded.
The answer is (x - y)(x + y)(x^4 + x^3y + x^2y^2 + xy^3 + y^4)(x^4 - x^3y + ...). Split as a difference of squares first, then use the fifth-power formulas.
Answer: (x-1)^2 (y-1)^2. Expanding everything gives a nine-term expression that is exactly the square of xy - x - y + 1, which itself factors as (x-1)(y-1).
The answer is ab = -1/4, from a = -1/2 and b = 1/2. The solution set lying between the roots forces a < 0, and the constant term 1 pins down the scale factor.
A linear equation in two variables needs both exponents equal to 1, giving 2a - b = 2 and 3a + 2b = 17, whose solution is a = 3, b = 4.
The answer is (a, b) = (4, −10). The coefficients fix the root sums at 5 and −3, so the intervals must be (1, 4) and (−5, 2), whose intersection is (1, 2).
a = -5, b = 25, so a - b = -30. The interval's centre 1 must equal -5/a and its radius 5 must equal b/|a|, which pins both parameters at once.
The answer is a ≤ 1. Substituting x = 3 gives 2a, and "3 not in A" means 2a > 2 must be false, so 2a ≤ 2 — no sign chart is needed anywhere.
Substituting the origin into sqrt((x-1)^2 + y^2) x (x - a) = 1 gives 1 x (-a) = 1, so a = -1.
(f - g)(x) = 4x + 8. Subtract the whole of g, so the minus sign must reach the -6 and turn it into +6, then combine 5x - x and 2 + 6.
f(2) = 18. Writing (x + 1)f(x) = (x^3 + 2)Q(x) + x + 2 with Q linear, the root x = -1 and f(0) = 4 pin down Q(x) = 2x + 1.
k = 3. The function is point-symmetric about (5, 7) and the bounds 1 and 13 are symmetric about 7, so the interval k − 9 ≤ x ≤ 3k + 7 must be centred at x = 5.
k = 5. The inequality is symmetric about x = 4, so the interval midpoint (k - 8 + 2k + 1)/2 must equal 4, giving 3k - 7 = 8 and the solution set -3 < x < 11.
Simplify each surd to a multiple of sqrt 3: -1, -1/3, +1/3 and +6. The two middle terms cancel exactly, leaving 5 sqrt 3.
The answer is 2022 ≤ λ ≤ 2026. Only the square root constrains λ: its radicand needs (λ − 2024)² ≤ 4, which is |λ − 2024| ≤ 2 centred on 2024.
m = 13. Substituting x = 2 turns 2mx into 4m, so the equation becomes 36 + 4m = 88; subtracting 36 then leaves 4m = 52, and dividing by 4 gives m = 13.
The answer is 5/4. Divide to isolate the conjugate of z-1, rationalise the denominator, conjugate both sides to recover z, then use |z|^2 = a^2 + b^2.
N = 42u²v − 25uv² + 3v³. Add the subtracted trinomial to both sides, then combine like terms carefully — the uv² terms are −10 and −15, giving −25.
P = 12m²n + 3mn² + 27mn. Because P is the one being subtracted, it equals the first polynomial minus the second, so every sign in the second one flips.
Answer: |Z| = 2. Write Z = r e^(i theta); the common factor e^(-i theta) has modulus 1, leaving root(r^2 + 16/r^2) = 2 root 2 and the perfect square (r^2 - 4)^2 = 0.
Answer: (-infinity, -3), with b = 6. Since (2, +infinity) is a solution interval, x = 2 is a root, giving b = 6 and the factorisation (x+3)(x-2) > 0.
The reciprocal of 1/(5y²) is 5y², valid for y ≠ 0. Flipping a fraction swaps numerator and denominator; the check is that the product of a value and its reciprocal is 1.
Adding all three equations gives 2(t1+t2+t3) = 24.62, so the total is 12.31 s. Subtracting each given pair sum yields t1 = 5.59, t2 = 2.97 and t3 = 3.75 s.
Answer: y = 2254 and x = 2231.46. Clear both denominators to get x = 0.99y and x = 0.98(y+23), equate them, and a coefficient of only 0.01 is left over.
Adding the two equations gives 5x + 5y = 29, so x + y = 29/5 in one line - no need to find x and y separately.
Solving 2018 - 5(n - 1) < 0 gives n > 404.6, so the first negative term is u_405 = -2. The previous term u_404 = 3 is still positive, confirming the cutoff.
Answer: divide 8x^3 by 2x, giving 4x^2. Polynomial long division always starts by dividing the leading term of the dividend by the leading term of the divisor.
Solving 3 + 7(n - 1) > 2018 gives n > 288.86, so n = 289 and u_289 = 2019. The previous term u_288 = 2012 is still below 2018, pinning the answer.
The solution is x = 126/5, y = 18/5, z = −24/5, w = 3/5, m = −41/5. Using y = 6w and x = 30 + z reduces everything to z, and the second equation pins w = 1 + z/12.
The solution is x = 17, y = 20, z = −13, w = −13/3. Clearing the fractions early and chaining the substitutions x = 30 + z and w = z/3 reduces everything to one equation in z.
The solution is x = −71, y = 68, z = −101, w = −709/6. The first equation gives x + y = −3, and x = 30 + z turns it into y + z = −33, leaving a single equation in y and z.
Answer: x = 17, y = 20, z = −13, w = −125/6. Clearing the first equation gives x − y = −3, and matching the second and third equations makes the z-terms cancel outright.
The gradient is m = 3. Add 6x to both sides to get 2y = 6x + 5, divide every term by 2 to reach y = 3x + 5/2, and read the coefficient of x as the slope.
The graph is an upward parabola with vertex (0, 1) and axis x = 0 - the parent y = x^2 shifted up one unit. Its range is y >= 1, so it never meets the x-axis.
The vertex is (−5, −4) and the V opens downward with slopes ±1, maximum −4, y-intercept −9, and no x-intercepts since every output is at most −4.
The solution is the filled diamond with vertices (4,0), (0,3), (-4,0) and (0,-3). Intercepts plus double reflection symmetry give the whole region quickly.
The solution is the closed upper half-plane. Because y >= 0 has no x in it, the boundary is the x-axis itself and the shading covers everything at or above it.
The solution is the unbounded wedge x ≥ 2, y ≤ 6 − x with corner (2, 4): right of the line x = 2 and on or below y = 6 − x, both boundaries included.
The height is x - 3 + (7x - 9)/(x^2 - 2). Long division leaves a nonzero remainder 7x - 9, so the height is not a polynomial - checked at x = 2, where it equals 1.5.
The hottest spice rates 26.5. Doubling each equation clears the halves, adding all three gives x + y + z = 43.5, and subtracting recovers each rating in one step.
Answer: for k not equal to 1 only (0,0,0); for k = 1 the whole line (0,t,t). Eliminating between equations 1 and 3 forces x = 0, then (k-1)z = 0 splits the cases.
Answer: (x, y, z) = (0, t, t) for any real t. The determinant is zero, so besides the trivial solution the system has a whole line of them through the origin.
The daily distances form an arithmetic progression, so S = (a1 + an)/2 x n. With S = 150 and a1 + an = 10 this gives 150 = 5n and n = 30 days.
The APs 2, 5, 8, ... and -1, 6, 13, ... (1000 terms each) share terms with x = 2 mod 3 and x = 6 mod 7, i.e. x = 20 mod 21 up to 2999, giving 142 values.
The graph of g is the graph of f translated 9 units straight down, moving the vertex from (0, 0) to (0, -9). The V shape and both slopes stay unchanged.
At the origin both z = 0 and z = 2 satisfy the equation, so no single-valued f exists. The surface is x^2 + y^2 = sqrt(8z) - z^2 for 0 <= z <= 2.
The solution is h > 1.3397214. No h ≤ 0 works; for h > 0 the inequality is equivalent to h⁴ − 1.8h − 0.81 > 0, whose single positive root is found by Newton's method.
The answer is x ≤ 1. Multiplying through by the LCD 6 removes both fractions, and dividing by −11 at the very end is the step that flips the inequality sign around.
The solution is x < 2 with x ≠ −1. Factoring gives (x + 1)²(x² − x + 1)(2 − x) > 0, and since the square and the quadratic are never negative, only 2 − x sets the sign.
A 1001-term AP from 1 to 2018 has d = 2017/1000. The 501st term is 1 + 500d = 1 + 2017/2 = 2019/2 = 1009.5, exactly the midpoint of 1 and 2018.
Placing 3 numbers between 2 and 22 makes a 5-term AP, so 22 = 2 + 4d and d = 5. The inserted means are 7, 12 and 17, giving 2, 7, 12, 17, 22.
The answer is x > 3. The branch x < −2 is wiped out by x ≥ −1 because the two conditions cannot both hold, leaving only the x > 3 branch.
Answer: (0.4433, 0.7268) and (-0.6025, 4.9100). Setting the two expressions equal and clearing the denominator gives the quadratic 9.36x^2 + 1.49x - 2.5 = 0.
Answer: about (0.4569, 0.6724) and (-0.9500, 6.2998). Clear the denominator to get 288x^2 + 142x - 125 = 0, solve, then check neither root hits the pole.
The intersection is the degenerate interval {−10}. Both intervals include −10 as a closed endpoint, and nothing else lies in both, so the overlap is one point.
Answer: t inverse(x) = x^2/9 - 6 for x >= 0. Swap x and y, divide by 3 before squaring, and the restriction comes from the original range being non-negative.
No, it is wrong: a comma in set-builder notation means AND, so the set is empty. To describe [2, 5] ∪ {10} you must write the word or, as in {x | 2 ≤ x ≤ 5 or x = 10}.
It is neither. f(-x) = -x sqrt(9 - x), which matches neither f(x) nor -f(x), and the domain x >= -9 is not symmetric about zero to begin with.
There are four solutions, including the exact pair (3/5, 4/5, -5) and (-1, 0, 1); the other two involve sqrt 11. Every candidate must satisfy x^2 + y^2 = 1.
The integers are 23, 24 and 25, so the largest is 25. Let the first be x, write the sum as 3x + 3 = 72, solve x = 23, then add 2 for the largest.
x = 6. When y = 2x^2 - 21x + 64 and y = 3x + a meet exactly once the quadratic has a double root, so x = 24/4 = 6 and a = -8. Full discriminant walkthrough.
The line is y = −4x + 3. Use point-slope form, watch the double negative x − (−1) become x + 1, then convert to slope-intercept form and check the point.
The equation is y = (2/3)x + 9. Start from point-slope form y − 5 = (2/3)(x + 6); the fraction clears cleanly because the x-coordinate −6 is divisible by 3.
x ≈ 128.0002, and the intended exact answer is 128. Both decimals are rounded versions of 3/110 and 384/110, whose quotient is exactly 128.
The solution is x = -3. Every term already has denominator 3, so each side collapses to a single fraction and the equation reduces to x - 1 = 2x + 2.
The equation is 50x + 20y = 2000, or 5x + 2y = 200 after dividing by 10. One valid pair is x = 20 adults and y = 50 children, since 1000 + 1000 = 2000.
Answer: all real numbers. Both sides expand to 3 - 8x, so the equation reduces to 3 - 8x = 3 - 8x, an identity rather than a condition that pins down x.
The solution is u = −1127/300 ≈ −3.7567. Reduce 4520/1200 to 113/30, write 0.01 as 1/100, then subtract over the common denominator 300.
Answer: x = -209/7, about -29.857. Combine each side into a mixed number, subtract to isolate the coefficient 7/45, and let the common denominator 45 cancel.
The equation has no solution. Clearing the nested brackets gives 6x - 23 = 6x - 2; the x terms cancel and leave the false statement -23 = -2.
The solution is x = 7250 exactly. Distribute 0.03 to get 0.024x − 150, flip the sign of the whole bracket, collect to 0.976x + 150 = 7226, then divide.
The solution is x = 442625/61 ≈ 7256.15. The same expansion gives 0.976x + 150 = 7232, but 7082/0.976 does not terminate, so the exact answer is a fraction.
Answer: x = 1,646,872.17, exactly 2653438800000/1611199. Factor x out of the left side, combine the bracket into one fraction, then multiply by its reciprocal.
The solution is x = 29300000/171 ≈ 171345.03. Multiply 0.0005 by 52 to get 0.026, collect 1x + 0.026x = 1.026x, then divide 175800 by 1.026.
Answer: x > 80/3, about 26.67. Subtract x to get 0.3x - 5 > 3, add 5, then divide by the small positive coefficient 0.3 - not by 2 as for 3x - 5 > x + 3.
The answer is n >= -23/6. Distribute the -2, combine constants to 29, add 8n to both sides, then divide by the positive 6 so the sign never flips.
The answer is n >= -5. Distributing -3 gives 31 - 3n, and after adding 8n the constants land on a whole number, so the bound is an integer, not a fraction.
Answer: x > -543/13, about -41.77. Simplify each side to 3x - 299 and 8.2x - 81.8, then divide by the positive 5.2 so the inequality sign is unchanged.
Answer: x > 10. Multiply by the LCD 20 to reach 5(x+2) - 4x > 20, expand to x + 10 > 20, and subtract; the x terms nearly cancel, leaving a single step.
Answer: x > 6. The trap is 4 - (x-2)/2 = 5 - x/2, not 3 - x/2. Collect the x terms over the denominator 6 to reach 5x/6 > 5 and finish in one division.
Answer: x < 11. Multiply both sides of (2x-1)/3 > (3x-5)/4 by the positive LCD 12, distribute to reach 8x-4 > 9x-15, then isolate x; no sign flip is needed here.
The answer is the empty set. Subtracting x from both sides erases the variable and leaves -1/3 > 4, a false statement, so no value of x can ever work.
The solution is x <= -1/2. Combine each side over a common denominator, multiply through by 12 to clear fractions, then isolate x from 14x - 2 <= 12x - 3.
Answer: x > -1. Multiply by the positive LCD 10 to get 5(x-1) < 2(7x+2), expand to 5x-5 < 14x+4, and move the x terms so the coefficient stays positive.
log y = b log x + log a. Because the unknown is the power, both axes need logs: gradient b and intercept log a come straight off a log-log graph.
Multiplying through by sqrt(x) gives y*sqrt(x) = px + q, so plotting Y = y*sqrt(x) against X = x is a straight line of gradient p and intercept q.
log y = -(log p)x + log q. Plotting log y against x (not log x) gives a straight line of gradient -log p and intercept log q.
1/y = (b/a)(1/x) - 1/a. No logarithm is needed here: the variables are tangled in a product, so reciprocals - not logs - straighten the graph.
The value is exactly 1. Because 18 = 3^2 * 2, the product rule gives log_3(18) = 2 + log_3(2), so the numerator and denominator are literally the same number.
The equation has the single root x = -3, so the sum of its roots is -3. Change of base turns the quotient of logs into one logarithm with base x + 5.
The solution is x > 16, so the least positive integer is 17. Merge the right side into one logarithm, compare arguments since base 5 > 1, and the x terms cancel.
The solution is x ≤ −5/2, x = 0, or 1/6 ≤ x < 1/2. Split on the sign of x, divide by x (flipping the sign when x < 0), and use that log base 1/3 is decreasing.
The quotient is 3x + 2 with remainder 3, so the result is 3x + 2 + 3/(x - 2). The nonzero remainder equals f(2) = 3, exactly as the Remainder Theorem predicts.
The quotient is 2x - 4 with remainder 2, i.e. 2x - 4 + 2/(3x + 2). Because the divisor leads with 3x rather than x, each quotient term needs a division by 3.
The quotient is 4x^2 + 7x + 15 with remainder 78, i.e. 4x^2 + 7x + 15 + 78/(2x - 5). The divisor is non-monic, so every quotient term comes from dividing by 2x.
The quotient is x^2 + 4x - 5 with remainder 0, so x + 3 is a factor and the cubic factors completely as (x + 3)(x + 5)(x - 1) with roots -3, -5 and 1.
The answer is h = 4p^2 - 20pk. Multiplying both sides by 4p frees h, then 4p is distributed over both terms of p - 5k, giving two terms and not one.
Proof that if a square matrix commutes with all matrices of the same size, it must equal λI. Uses matrix units E_ij to force every off-diagonal entry to zero.
AB is the 3x2 matrix with rows (11,11), (4,4) and (20,20). A^T B^T cannot be formed because a 3x3 cannot multiply a 2x3; the valid transpose identity is (AB)^T = B^T A^T.
The answer is x = (36 - sqrt 71)/12. Cross-multiplying gives 6x = a^2, and squaring the surd produces 72 - 2 sqrt 71 before dividing by 4 and then 6.
The minimum is (4√2 − 3)/6 ≈ 0.4428, reached at a = (5 − 3√2)/2 and b = 3√2 − 4. Substituting b = 1 − 2a and setting the derivative to zero gives (1 + 2a)² = 8(1 − a)².
The minimum value is 3/16, reached when a = b = c. A power-mean step plus a two-term lemma pins the sum of squares below by 3/4, which is exactly enough.
Answer: a <= 7. The sum of four absolute values has minimum 7, attained on the whole middle interval [-1, 1] between the two inner points, not on [-3, 1].
Read the expression as the total distance from x to -1 and to 2. That sum is smallest for any x between them, where it equals the gap 2 - (-1) = 3.
Mixing 10 L of the 20% solution with 10 L of the 50% solution gives 20 L at 35% alcohol. Set up the volume and pure-alcohol equations and solve the system.
Answer: 25. Modulus rules collapse the expression to |Z2|^2, so only Z2 = (7i^88 + 24i^33)/(5i) matters; it simplifies to 24/5 - 7i/5 with |Z2| = 5.
The value is 27. Cancelling Z2 reduces the expression to |Z1|^3, and the two given equations force conjugate Z1 = -3i, so Z1 = 3i and |Z1| = 3.
The product is 21x^4 + 28x^3y + 62x^2y^2 + 36xy^3 + 45y^4. Sorting the six partial products by total degree in x and y shows only the x^2y^2 pair combines.
(4x + 1)(3x + 2) equals 12x² + 11x + 2. FOIL multiplies the First, Outer, Inner and Last pairs, then the two middle terms 8x and 3x combine into 11x.
The product is 10x^4 + 24x^3 - 63x^2 + 27x. Distribute 5x^2 and then -3x across the trinomial, then combine the two like x^3 terms and the two like x^2 terms.
(6x² − 7)(5x² − 6) equals 30x⁴ − 71x² + 42. Treat x² as a single unit, multiply the four pairs of terms, then combine −36x² and −35x² into the middle term −71x².
The answer is 85, a real number. The factors are complex conjugates, so the product collapses to 9^2 - (2i)^2 = 81 + 4 and the i terms cancel exactly.
The product is 4x^4 + 21x^3 + 7x^2 + 16x + 30. Distribute x over all four terms, then distribute 5, then add the two rows by matching powers of x.
The product is 56y^6 + 21y^5 - 21y^4 - 56x^2y^3 - 35y^3 + 21y^2 - 56xy^2. Multiply coefficients and add exponents on y for each of the seven terms.
The product equals 2 times the cube root of 3. Combine the two radicals into the cube root of 24, then pull out the perfect cube factor 8 that divides 24.
The product is -18x^4 + 8x^3 + 59x^2 - 28x + 14. The first factor has no x term, so only the x^2 column receives contributions from both halves of the distribution.
The product equals 2x² − 4x + 1. Group the x-coefficients into the conjugate pair 2 ± √2, use (2−√2)(2+√2) = 2 for the x² term, and the surds cancel in the x term.
The product simplifies to (x - 3)/(x + 8). Factor all four quadratics, cancel the shared (x - 6), (x + 3) and (x - 8) factors, and never cancel across a sum.
The product is x^3 - 9x^2 + 14x + 24. Distribute x and then -4 across the trinomial, and note the constant comes out positive because -4 times -6 is +24.
The product is x³ + 1. Distributing gives x³ − x² + x + x² − x + 1, where four terms cancel — the shortcut is the identity a³ + b³ = (a + b)(a² − ab + b²) with a = x, b = 1.
The product is x^3 + 1000. The second factor is exactly a^2 - ab + b^2 with a = x and b = 10, so the identity collapses six terms down to two.
The product is 2x^2 + 9xy - 5y^2. Distributing gives four terms, and the two middle terms -xy and +10xy combine into the single like term 9xy.
The product is (x-1)/(x+2). Factor all four quadratics first, then cancel one (x-1), one (x+3) and one x across the two fractions before multiplying anything out.
The product is 2x^4 + 6x^2y^2 + 6x^2z^2 + 4y^4 + 6y^2z^2. Only the x^2y^2 terms combine; the other four monomials involve different variable patterns and stay separate.
Answer: no real solution. A sum of nonnegative squares is zero only if both vanish, and 1.6 does not equal 1.5; expanding confirms a discriminant of -0.08.
The answer is 15 pens. Set the price as 240/x, write the condition 240/x − 240/(x+1) = 1, clear the denominators to x² + x − 240 = 0, and discard the negative root.
Rows of 1, 2, 3, ... trees give the triangular sum n(n + 1)/2 = 3003, so n^2 + n - 6006 = 0 and n = 77. Checking, 77(78)/2 = 3003 exactly.
The answer splits into seven cases. The sign of a decides whether the solution is a bounded interval, two rays, all of R or empty, and b decides the boundary.
The answer splits on a: all reals if a > 1, all x ≠ −1 if a = 1, x > −1/2 if a = 0, and intervals built from (−1 ± √(1−a))/a otherwise.
After moving 2x across, the inequality factors as (x−2)(ax−1) ≥ 0. The answer splits at a = 1/2: outside the roots 2 and 1/a, with all of ℝ when the roots coincide.
The discriminant is the perfect square (3a + 4)², so the roots are always −3 and 4/a. The solution set splits by the sign of a, with a = 0 and a = −4/3 as separate cases.
The decomposition is 1/n - 2/(2n + 1). Factoring the denominator as n(2n + 1) gives two distinct linear factors, and comparing coefficients yields A = 1, B = -2.
The result is 2/(x - 1) + 1/(x - 1)^2. A repeated linear factor needs one term per power, and matching coefficients gives A = 2, B = 1.
The decomposition is 1/(x - 1) + 2/(x - 1)^2. A repeated linear factor needs one term per power, and equating coefficients gives A = 1 and B = 2.
The factorization is (3x − 2y)². Both end terms are perfect squares, 9x² = (3x)² and 4y² = (2y)², and the middle term −12xy is exactly −2(3x)(2y).
x^2 - 8xy + 16y^2 = (x - 4y)^2. Check that 16y^2 is (4y)^2 and that the middle term equals -2 times x times 4y, then read off the square of a difference.
The perimeter is 43 m. Setting the width to x gives x(2x+5)=88, a quadratic whose positive root is 5.5 m, so the length is 16 m.
The perimeter is 40 feet. Adding the three expressions gives 3x^2 + 3x - 20, which at x = 4 evaluates to 40 - combining like terms first is faster.
The difference simplifies to 12a^2b^2 - 5ab^5, a binomial of degree 6. See why subtracting cancels the a^3b terms and why degree comes from adding exponents.
The missing quotient entry is A = −3, so the quotient is x − 3 with remainder 0. Learn how each cell of a polynomial division table is generated, step by step.
The quotient is x − 2 and the remainder is 8x − 5. A quadratic divisor rules synthetic division out, so long division is the method that applies here.
The quotient is x and the remainder is −6x + 5, so the cubic equals x(x² + 2) − 6x + 5. Both the dividend and divisor have gaps that need placeholder zeros.
The positive value is x = 30. Cross-multiplying gives x^2 = 900, so x is the geometric mean of 20 and 45, and both ratios then reduce to 2 : 3.
a + b + c equals y(1) = 64a - 14, and a must be positive, so any value greater than -14 works. Evaluating at x = 1 is the trick that avoids solving for b and c.
Printer B prints 4200 pages. Printer A's rate of 360 pages per hour means the job is 2520 pages, so B works at 840 pages per hour and produces 4200 in 5 hours.
The product AB has rows (0, 6, 16), (-7, -2, -3) and (-2, 2, 6). Each entry is the dot product of a row of A with a column of B, so nine dot products are needed.
Expand (x + y)^3 = x^3 + 3x^2y + 3xy^2 + y^3 and 3xy(x + y) = 3x^2y + 3xy^2; subtracting cancels both mixed terms and leaves exactly x^3 + y^3.
The identity is true for all sets. Rewrite the symmetric difference, note its complement is (A∩B) ∪ complement(A∪B), and distribute the removal of (A∩B∩C) over that union.
Completing the square turns the factors into 2(x-1)^2 + 1 and (x-1)^2 + 1, each at least 1. Their product can equal 1 only when both equal 1, forcing x = 1.
Both factors complete the square about x = 2 into (x-2)^2 + 1 and 2(x-2)^2 + 1, each at least 1, so their product is 1 only when x = 2.
The discriminant is a^2 - 4a + 8 = (a - 2)^2 + 4, which is at least 4 for every real a. A strictly positive discriminant guarantees two different real roots.
Answer: x = 20.568376 or x = 2.190902. The discriminant is b^2 - 4ac = 3.1777162025, and dividing by 2a = 0.194 magnifies any early rounding tenfold.
Answer: x = 21.675719 or x = 2.333560. Keep every decimal in the discriminant b^2 - 4ac = 3.52008685 instead of rounding, then divide by 2a = 0.194.
The roots are x ≈ 142.379 and x ≈ 17.621, exactly (1360 ± 105√102)/17. The discriminant is 426147.84 − 167051.52 = 259096.32, whose square root is 509.01505.
The roots are x = (11 - sqrt29)/2 = 2.807 and x = (3 + sqrt13)/2 = 3.303. Split at x = 3, solve a quadratic on each branch, and keep the roots that fit.
Answer: 2 − 2√3 ≤ x ≤ 2 + 2√3. The roots give f(x) = a(x + 1)(x − 5), the minimum −18 at the vertex x = 2 forces a = 2, and then you solve 2x² − 8x − 10 ≤ 6.
The real solutions are x = −12 and x = −13. Substitute u = x + 4 to turn a quadratic-in-form equation into u² + 17u + 72 = 0, factor, then convert back.
The answer is -1/2 <= x <= 2/3. The hidden leading coefficient is -6, so the product is positive between the roots rather than outside them.
The answer is 3/2 < x < 2. The hidden leading coefficient is -2, so the product is positive strictly between the roots rather than outside them.
Answer: x < -1/2 or x > 2. Factor the non-monic quadratic as (2x + 1)(x - 2), find the critical points -1/2 and 2, and keep the outside intervals.
The answer is x <= -1/3 or x >= 1. The AC method splits -2x into -3x + x, factoring it to (3x + 1)(x - 1), and the parabola opens upward from there.
The only answer is x = 1/2. The trinomial is the perfect square (2x - 1)^2, which is never negative, so equality is the only way to satisfy the inequality.
Answer: no solution. The quadratic is the perfect square (2x - 1)^2, and a real square is never negative, so the strict inequality has an empty solution set.
The answer uses the roots (-8 plus or minus 8 root 10)/3, about 5.766 and -11.099. The trinomial does not factor over the integers, so x = 6 is not a root.
The answer is m < -6 or m > 2. Squaring -m removes the sign, and distributing -4 flips both inner signs, giving the factorable m^2 + 4m - 12.
There is no solution. Expanding and rearranging gives 3x^2 - 6x + 7 < 0, but the discriminant is -48, so this upward parabola stays strictly above the axis.
Answer: [-1, 0] union [2/3, 4]. Factor each quadratic, solve the two inequalities separately, then intersect the solution sets on one number line to finish.
Answer: [-1, 0] union [5/3, 4]. Multiply the first inequality by -1 to flip its sign, factor 3x^2-5x as x(3x-5), then intersect the two solution sets.
Answer: x < -2*sqrt(11363)/11 or x > 2*sqrt(11363)/11, about +/-19.3813. Evaluate the right side to 19285, divide by 55, and keep the bound as a fraction.
It factors as (x-1)(ax+1), so the answer needs five cases - not three. The a < 0 branch splits further at a = -1, where -1/a equals 1 and the roots coincide.
The solution set is (−∞, 1) ∪ (3, ∞). The roots 1 and 3 split the line into three intervals, and the product of two linear factors is positive only outside the roots.
The answer is x < -1 or x > 3. The parabola opens upward, so the product is positive outside the two roots and negative in the strip between them.
The solution is x < -1 or x > 3. Subtract 3 to reach (x - 3)(x + 1) > 0, then note that an upward-opening parabola is positive outside its two roots.
The answer is x < 1 - root 2 or x > 1 + root 2. The trinomial does not factor over the integers, so the quadratic formula supplies the two boundary points.
The answer is the open interval (-1, 3). The trinomial factors as (x - 3)(x + 1), and an upward parabola is negative only between its two roots.
The solution is x < -1 or x > 5. Factoring gives (x - 5)(x + 1) > 0, and testing one point in each of the three intervals shows where the product is positive.
The answer is -1 < x < 5. Factor to (x - 5)(x + 1), then use the upward-opening parabola: a positive quadratic is negative only strictly between its two roots.
The answer is all real x. The discriminant is -3, so there are no real roots, and completing the square gives (x + 1/2)^2 + 3/4, a positive minimum of 3/4.
The greatest possible value of n is 14. Factoring f(x) = a(x - 1)(x + 7) gives b = 6a, so a + b = 7a, and the smallest allowed integer a is 2, giving 14.
The answer is 7/4 < k < 13/6. Sign conditions f(0) > 0, f(1) < 0, f(2) < 0, f(3) > 0 pin the parameter down, and the product of the roots gives a check.
All four roots are real: m ≈ −0.86460, 3.37180, 11.10473, 13.31466. A sign scan finds every bracket, and Vieta's sum and product confirm none were missed.
Answer: no real roots. A sum-of-squares split, 3(x^2+3x+3/2)^2 + 2x^2 + 5x + 21/4, shows the quartic never drops below 17/8, so all four roots are complex.
The roots are u = −1, u = 1 and the double root u = 3. Matching coefficients gives (u² − 1)(u² − 6u + 9), which factors further as (u − 1)(u + 1)(u − 3)².
Answer: x = 0.0115484893 and x = -0.0115485279. The quartic term dominates, so the roots are nearly the fourth roots of 1210.23/6.804e10 and nearly symmetric.
Answer: x = 947.0356 and x = -1332.1428. A tiny leading coefficient is not a negligible term at large x: at x = 1729 the quartic piece is already about 6079.
Answer: x = 0.9 exactly, plus x = -63.548577, 0.897443, 79.551134. Both sides hide the factor (0.9 - x), so the quartic drops to a cubic instead of expanding.
The real roots are x ≈ 8.4610649 and x ≈ −0.3736462, plus a complex pair. This quartic has no rational root and no integer quadratic split, so bracket and use Newton.
The roots are x = 1, x = 7 and x = 4 ± √21. Both quadratic factors share x² − 8x, so setting t = x² − 8x turns the quartic into t² + 2t − 35 = (t + 7)(t − 5).
Answer: x = 205/18, about 11.3889. Divide by 3 before squaring, so root(2x - 1) = 14/3; then 2x - 1 = 196/9, giving 2x = 205/9 and finally x = 205/18.
The only solution is x = 3. Squaring produces x = 0 and x = 3, but x = 0 makes the right side negative, so it is extraneous and must be discarded.
Answer: x = 7/2 or x = -3/2. The radicand factors as 4(x - 1)^2, so the root is 2|x - 1|; the absolute value produces exactly two solutions.
Answer: x = 2. Squaring cancels the x^2 terms and leaves -3 = -2x + 1, so x = 2; it satisfies the domain condition x >= 1, so it is not extraneous.
Answer: no real solution. Squaring gives 3x^2 - 14x + 24 = 0, whose discriminant is 196 - 288 = -92, so no real x satisfies the squared equation either.
Answer: no solution. Squaring gives x^2 - 6x + 25 = x^2 - 6x + 9, so all the x-terms cancel and the equation reduces to the false statement 25 = 9.
Answer: x = 0 or x = 6. Squaring is safe because the right side 5 is positive; the constant 25 then cancels from both sides, leaving x(x - 6) = 0.
The only solution is x = 4. Squaring produces x² − 4x = 0 with roots 0 and 4, but x = 0 makes the right side negative and is rejected.
There are two solutions, x = 0.9 and x ≈ 0.8974432593. Substituting u = 8.9 − x turns the equation into 71(u−8) = (8.9−u)√((u−8)(u+8)), exposing the boundary root u = 8.
The result is m^8/(256 n^44). Each factor gets exponent -4: (-4)^(-4) = 1/256, m^(-2) becomes m^8, and n^11 becomes n^(-44), which drops to the denominator.
The range is 0 <= z <= 22. Writing 6x - 8y as 4(x + y) + 2(x - 6y) lets you bound z directly from the two given intervals instead of solving for x and y.
A column-space basis gives A = UC = sum of r outer products; subadditivity of rank shows fewer than r is impossible. Two halves, two different tools.
The ratio is x : y = 1 : 2. Clearing the /100 and then the 2/3 turns the sentence into 12x + 15y = 10x + 16y, which collapses to 2x = y.
The solution is (1/4, 4]. Split the chain into two rational inequalities, move everything to one side of each, and intersect the sign-chart results while excluding x = -1.
x = 5/3. The common denominator (x - 2)(x + 2) is exactly x^2 - 4, which reduces the equation to 3x - 2 = 3, and 5/3 avoids the excluded values 2 and -2.
The solution is X = 21/47 ≈ 0.4468. Combine 450X − 6X into 444X, cross-multiply to 3150 + 54X = 7104X, then reduce 3150/7050 by its gcd 150.
Answer: x = (-225 +/- 25*sqrt(161))/2, about 46.107 and -271.107. The numerator collapses to the constant 2500, leaving x^2 + 225x - 12500 = 0 to solve.
The numerator collapses to -2500, giving x^2+225x+37500=0 with discriminant -99375, so no real solution exists: x = (-225 +/- 25i*sqrt(159))/2 over C.
There is no real solution. Combining the fractions collapses the numerator to the constant −2500, so the equation becomes x² + 225x + 37500 = 0 with discriminant −99375 < 0.
The answer is [2/3, 1). Move the 1 across and combine over one denominator to get (3x - 2)/(1 - x) >= 0 - and never cross-multiply by 1 - x.
The answer is the half-open interval [1/2, 1). The numerator's zero is included because 0 satisfies >=, but x = 1 is excluded since it kills the denominator.
Answer: [-1, 1/2]. The denominator has discriminant -7 so it is positive for every x, meaning the sign of the whole fraction is decided by the numerator alone.
The answer is the open interval (-1/2, 4). A quotient is negative exactly when numerator and denominator have opposite signs, which happens only between the roots.
The answer is -1/2 < x <= 2/3. Subtracting 1 gives (2 - 3x)/(2x + 1) >= 0, and the left end stays open because the denominator vanishes exactly there.
The answer is x < (-3-root 37)/2 or 1 < x < (-3+root 37)/2. Never cross-multiply: the sign of x - 1 is unknown, so combine into one fraction and use a sign chart.
Answer: 1 < x <= 2, the interval (1, 2]. Move everything to one side, combine into (6 - 3x)/(x - 1) >= 0, and use a sign chart - never multiply by x - 1 of unknown sign.
The solution is the interval (2, 5). Subtracting 1 and combining into (5 - x)/(x - 2) > 0 avoids multiplying by x - 2, whose sign is unknown.
The solution is -1/2 <= x < 0 together with the single point x = 1/3. The squared factor never changes sign but does contribute one isolated equality point.
Answer: x in ((-1 - root409)/6, -2) union ((-1 + root409)/6, infinity). Combine the polynomial parts into (3x - 5)/6, then sign-chart (3x^2 + x - 34)/(x + 2).
The solution is (-inf, 1] union (2, 3). Mark the three critical points, test each interval, and keep x = 1 closed while excluding x = 2 and x = 3.
The answer is (-3, 1]. Both parts share the factor x - 3, which cancels but still removes x = 3 from the domain - a hole, not a sign change.
Multiply top and bottom by the conjugate root3 - root2; the denominator becomes 3 - 2 = 1, so the answer is just the expanded numerator, 3 sqrt 6 - 7.
The answer is (17√11 − 34)/7. Multiplying by the conjugate 2 − √11 turns the denominator into 4 − 11 = −7, so the minus sign then moves up into the numerator.
The answer is (2 root 7 - 2 root 2) / 5. Multiply top and bottom by the conjugate root 7 - root 2, so the difference of squares clears both radicals.
The cube root of 7/y simplifies to the cube root of 7y² divided by y. Multiply inside the radical by y²/y² so the denominator becomes the perfect cube y³.
The result is sqrt(11)/11. Multiply top and bottom by sqrt(11) so the denominator becomes 11; the numerator and denominator share no factor, so nothing cancels.
The result is 7 root 5 + 14. The conjugate gives a denominator of 5 - 4 = 1, so the fraction disappears entirely and the answer has no denominator at all.
The cubic has three real roots: -2.7875545, 3.2205249 and 4.5670296. No rational root exists, since the only candidates are the divisors of 41, namely 1 and 41.
The answer is x ≈ 4.9807×10⁻⁵, not exactly 5×10⁻⁵. The right side is just 1.17% of 8×10¹², which shifts x by about 0.39% — small, but very far from zero.
The root is x = 4. Two reciprocals are equal only when their denominators are equal, so 4x − 5 = 11, giving 4x = 16 and x = 4, which is inside the domain.
The result is f(x) = −|x + 2| − 1. Reflecting negates the entire function, turning the −1 into +1, and the later shift down 2 brings the constant back to −1.
The result is f(x) = −|x + 2| + 1 with vertex (−2, 1). Reflection sends the −3 to +3, and the shift down 2 leaves +1 — the constant does not simply stay negative.
The result is f(x) = −|x| − 6 with vertex (0, −6) and arms of slope ∓1. Every output is negative, so the graph lies entirely below the x-axis with maximum −6.
The region is the part of the closed disk of radius 3 lying on or below y = x + 1. The boundary line cuts the circle where x = (-1 plus or minus root 17)/2.
The remainder is 26x - 47. If f(x) leaves remainder 2x + 1 on division by 2x^2 - 3x + 4, cube that remainder and reduce it using the rule x^2 = (3x - 4)/2.
The remainder is 26x - 47. Because f = qD + (2x + 1), the cube is congruent to (2x + 1)^3 modulo D, so only that cube has to be reduced by long division.
Use the Remainder Theorem to divide x³ − 2 by x − 1 in one step, without long division. Includes why the remainder is a constant and not a polynomial.
Since 27 = 3^3, the power-of-a-power rule gives 27^(-2) = 3^(-6), so 3^8/27^(-2) becomes 3^8/3^(-6) = 3^14 = 4,782,969 once the exponents are subtracted.
Euler's formula splits each exponential into cosine and sine; the cosine parts add and the sine parts subtract, giving a clean real-plus-imaginary form.
The slope-intercept form is y = x/4 - 7/4. Distributing the 1/4 and combining 1/4 with -2 over the denominator 4 gives a slope of 1/4 and intercept -7/4.
The first three terms form the perfect square (x + 4y)², since 16y² = (4y)² and 8xy is twice x·4y. The −4z term cannot be absorbed without knowing z.
The roots are x ≈ −29.572 and x ≈ 51.859, more than 81 apart. A tiny leading coefficient makes the parabola almost flat, pushing the roots far from the vertex at x ≈ 11.14.
Answer: x = 2 and -2 in the reals, plus 2i and -2i over the complex numbers. Factor x^4 - 16 as (x^2-4)(x^2+4), then split x^2 - 4 into (x-2)(x+2).
The outbound speed is 50 km/h and the return speed 60 km/h. Convert 4 h 24 min to 4.4 h, write both times as distance over speed, and solve 220/v = 4.4.
The answer is 3037. The characteristic equation (r - 2)^2 = 0 gives a_n = (A + Bn)2^n, so a_2025 = 6,149,925 * 2^2025 and b/c = 6,149,925/2025 = 3037.
Term 19 is 91.66672, so its integer part is 91. The gaps halve and alternate in sign, and the sequence converges to (a1 + 2a2)/3 = 275/3, not to 2a2 - a1.
The real solutions are x ≈ ±1.23561. Clearing the decimal gives x⁶ + 25x⁴ + 25x² − 100 = 0, and y = x² turns it into a cubic whose only positive root is y ≈ 1.526734.
Four real roots: x ≈ 0.00096368, 0.14670508, 0.15345189, 0.22484227, plus a complex pair. Bracketing by sign changes is what makes the near-double roots separable.
Answer: x = 2.982928 and x = -3.152475. Expanding gives the sextic x^6+104x^4+40x^3-10000 = 0, which has no rational root, so bracket and bisect both sign changes.
Answer: the system is inconsistent - no solution exists. Substitution reduces it to three equations in y and z that give y = 25/2, z = -9/2, values which fail the remaining equation.
The inequality is equivalent to (k - x)(t - y) > 0, so k - x and t - y must share a sign. Grouping turns four scattered products into one factored condition.
The cubic is irreducible over the rationals. Only 1 and -1 are candidate roots and both fail, so no linear rational factor exists and grouping cannot work either.
No largest m exists: substituting y = 1 − 3x gives 1/x − 3 + 1/(1 − 3x), which tends to +∞ as x → 1/3⁻, so the expression is unbounded above and the bound m ≥ value fails.
Answer: [2, 5] union {10}. Even powers never change the sign, so only (x-2)^3 and (x-5)^3 matter; the isolated point x = 10 belongs because the product is zero.
The expression is positive on (-inf, 0) union (2/3, 1). The numerator flips at 2/3 and the denominator at 0 and 1, so only two of the four intervals qualify.
The answer is 0.6x + 0.2. This is the weighted-average form: mixing fraction x at 0.8 with the rest at 0.2 gives a result rising linearly in x.
The answer is 1/a. The numerator is the perfect square (1 − √(b/a))² = (√a − √b)²/a, so the shared factor (√a − √b)² cancels, leaving 1/a for a > 0, b ≥ 0, a ≠ b.
Answer: 6pi/(pi + 4). Both numerator and denominator carry the factor 32pi; pulling it out leaves 6pi over 4 + pi, which cannot be reduced further.
The answer is -x^2 y + 7xy. The -3 must distribute over BOTH terms, turning -xy into +3xy - the step most worked solutions get wrong, ending up with -xy instead of +7xy.
The quotient simplifies to 7y⁵/(4x³). Reduce 28/16 to 7/4, subtract the exponents to get x⁻³ and y⁵, then move the x⁻³ down into the denominator to finish.
The simplified expression is a^2*b - ab^2 + b^2 - 2b. The minus sign in front of the second bracket flips every sign inside before like terms are collected.
The simplified form is 3x³ + x² + 2x − 6. Only the two cubic terms are alike, so they add to 3x³ while the x², x and constant terms stay untouched.
The result is (x+3)/(x+1). Expand and collect the numerator into x^2 + x - 6, factor it as (x+3)(x-2), then cancel the common (x-2) against the denominator.
The result is sqrt(3)*cbrt(4)/4. Pull 2 out of sqrt12 and 3 out of cbrt54, then multiply by cbrt(4)/cbrt(4) because a cube root needs two extra factors.
The quotient is 3^5 = 243. Rewrite 27, 9 and 81 as powers of 3, add the exponents on the top and on the bottom, then subtract them: 22 - 17 = 5, so the value is 243.
Answer: 3a, valid for a not equal to 0. Split off the coefficient, subtract exponents so a^2/a = a^1, and note why the original expression needs a nonzero denominator.
The answer is 12x + 5. Distributing the 4 gives 12x + 20, and only the constants combine, since 20 and −15 are like terms while the 12x term has no partner.
The answer is 4d/a^9. Adding the exponents -10 and 1 gives a^-9, and a negative exponent means the factor belongs in the denominator instead.
The answer is 15 - 8i. The first radical is a real 15, while sqrt(-64) must first be rewritten as 8i using i = sqrt(-1) - never as a real -8.
The simplified form is 57x - 40. Distribute the 5 across both terms inside the bracket, then combine 45x with 12x - only the like terms may be added.
Distribute -4 over all three terms: -12 + 32x - 8x^2. Then combine: 5x^2 - 8x^2 = -3x^2, 2x + 32x = 34x and -1 - 12 = -13.
The answer is 3x + 12. Distribute the 6 across both terms in the bracket to get 6x + 12, then combine the like terms 6x and -3x into a single 3x.
The answer is 6(x-1)/(x(x^2+10x+1)). Factor 6 from the top, take out x on the bottom and expand inside to x^2+10x+1, which has no factor of x-1, so the fraction is already in lowest terms.
The result is 3(x - y)^2(x + y). Because y - x = -(x - y), the odd cube flips sign to -3(x - y)^3, after which 3(x - y)^2 is a common factor of both terms.
Answer: 77x^5 y^2. Multiply the constants to 77, then add exponents of like bases: three separate x factors and one x^2 give x^5, and the two y factors give y^2.
The simplified form is 13f + 12. Distributing gives 7f + 28 + 6f − 16, so the f terms add to 13f while the constants 28 and −16 combine to give 12.
The answer is just y. Distributing -3 gives -6y + 3z, so the z terms cancel completely and only 7y - 6y survives out of the whole expression.
The simplified form is 27y - 109. Clear the inner bracket first to get 8 - 9[13 - 3y], then distribute -9 across both terms to reach 27y - 109.
The simplified form is 52y - 11. Distribute 8 and 3 across their brackets to get 40y - 32 + 12y + 21, then combine the y-terms and the constants separately.
The expression collapses to x² − 16y² = (x − 4y)(x + 4y). The two xy terms cancel exactly, and that cancellation is what leaves a clean difference of squares.
The simplified form is 2a/b. Divide the coefficients 8/4 = 2, then subtract exponents: a^(2-1) = a stays on top while b^(1-2) = b^(-1) moves to the bottom.
The result is 8m^2 + 20mn + 8n^2 = 4(m + 2n)(2m + n). Writing 9(m+n)^2 as (3m+3n)^2 turns the whole expression into a plain difference of two squares.
Factor all four quadratics: (3y-1)(2y+1), (2y+1)^2, (3y-1)(y+1) and (2y+1)(y-4). Inverting the lower fraction lets every factor but (y-4)/(y+1) cancel.
The result is 2(x+1)(x^3 - 2x^2 - 3x - 2). Factoring x^2 - 9 lets x + 3 cancel, after which (x+1) is a common factor of all four terms.
The expression collapses to 8x² + 12x + 4 = 4(x + 1)(2x + 1). All the cubic terms cancel, so a degree-3 expression simplifies to a quadratic.
The bracket collapses to -8a/(a^2-4). Dividing by 12a^2/(4-a^2) means multiplying by (4-a^2)/(12a^2), and the two sign-flipped quadratics cancel to leave 2/(3a).
The bracket collapses to 4t, so the first fraction is 32t/(1-t^2)^2 and the expression becomes a single fraction over (1-t^2)^2(1+t^2)^2.
Answer: (x-y)^2 (x+y)(x+y-1). Because (y-x)^2 equals (x-y)^2, both terms share (x-y)^2(x+y), and pulling that out leaves the simple bracket (x+y) - 1.
The result is (a + 2b)^2. Expanding gives a^2 - 4ab + 4b^2, and adding 8ab turns the middle term from -4ab into +4ab, flipping the sign inside the square.
The whole expression simplifies to 1. The denominator is the perfect square (root a + root b)^2 and the numerator is a difference of squares in root a and root b.
The expression collapses to 2 - a - 2b. The inner bracket becomes (2a+b)/(a+b), and its reciprocal cancels both (a+b) and (2a+b) at a stroke.
The answer is (a - x^2)/x, valid for x not zero. Rewrite the lone x as x^2/x so both terms share the denominator, then subtract the numerators in one step.
Answer: 6 root3 x^(62/21) / (35 y^(3/2)). Convert every radical to a fractional exponent, gather the constants into 18/(35 root3), then add exponents base by base.
The answer is 1/(x^20 y^24 z^12). Subtract exponents inside the parentheses to get x^10 y^12 z^6, multiply each by -2, then move them to the denominator.
The value is 1/2. Factoring 2^8 from the numerator and 2^9 from the denominator leaves 7*2^8 over 7*2^9, so the sevens cancel and the powers give 2^-1.
The answer is p^8 / (m^9 n^10). Subtract the bottom exponent from the top for each letter, then move any negative result to the other side of the fraction bar.
The answer is m^8 / (256 n^44). Divide inside first to get -4m^-2 n^11, then raise each factor to -4; the negative base to an even power turns positive.
The result is (b² + b√b + 2b − 3√b + 1)/(b + √b − 1). Flip the negative exponent, set x = √b so every power is a polynomial, cancel x + 1, then add 2√b.
The answer is 1. Any nonzero quantity raised to the power 0 equals 1, so the whole fraction of negative exponents never has to be simplified at all here.
The whole expression collapses to a^4 b. Factor ab out of the first bracket, apply the sum-of-cubes identity, distribute, and the ab^4 terms cancel exactly.
The answer is ∛(ab)/(∛a + ∛b). Setting x = ∛a and y = ∛b turns the denominator into (x + y)²/(xy), so the quotient collapses to xy/(x + y) after cancelling x + y.
The quotient simplifies to 3x. Combine the two fifth roots into one radical, reduce 729x⁶/(3x) to 243x⁵, and recognise 243x⁵ as a perfect fifth power.
It becomes (L - R)(L + R)/(H + R), and nothing cancels unless H = L, in which case it reduces to L - R. Distinct letters can never cancel against each other.
The simplified result is 9x - 8, choice (A). Distribute -1/5 to get x - 2, distribute -2 to get 8x - 6, then combine like terms; checks out at x = 0, 1 and 3.
Answer: f(x) = -x - 289/x with x not equal to 0. Split the single fraction into two terms, cancel one factor of x, and state the domain restriction.
The simplified form is −80v + 40, or 40(1 − 2v). Distributing −10 gives −90v + 40, and adding the loose 10v leaves −80v with the constant unchanged.
The expression simplifies to −125b²⁴/a¹⁸. Reduce inside the parentheses first with the quotient rule, then cube the coefficient and multiply every exponent by 3.
The answer is -243 x^10 y^20, not the -27 x^10 y^20 often written. Raise every factor to the fifth power: (-3)^5 = -243, and multiply the exponents 2 and 4 by 5.
The simplified form is −9q + 11r. Two of the three groupings carry a negative multiplier, so every term inside them flips sign before like terms are collected.
The whole expression collapses to just a. Combine each bracket over a+b, flip the second one because of the -1 exponent, cancel, then subtract the trailing 3b.
Distribute sqrt 2 to get 2 sqrt 12 - sqrt 16 = 4 sqrt 3 - 4, then 3 sqrt 3 - (4 sqrt 3 - 4) = 4 - sqrt 3.
The result is 3x√7. Combining under one radical gives √(63x²), and 63 = 9 × 7 releases a factor of 3 while x² releases x.
After simplifying and rationalising, the expression is -sqrt 3 + 2 sqrt 5, so a = -1, b = 2 and a + b = 1.
The result is 2x^5 sqrt(6x). Combine the two radicals into one, divide inside to get 24x^11, then pull out the perfect square factors 4 and x^10.
The three pieces are 17, 7 and 24, so the whole expression is 0. Note sqrt((-7)^2) = 7, not -7, and (-2 sqrt 6)^2 = 24, not -24.
Write a - 1 as (sqrt a - 1)(sqrt a + 1); the common factor sqrt a + 1 cancels, leaving 1/(sqrt a - 1) for a >= 0 with a not equal to 1.
The simplified form is x sqrt(x). Split x^3 into x^2 times x, apply the rule sqrt(ab) = sqrt(a) sqrt(b), then use sqrt(x^2) = x because x is nonnegative.
The expression collapses to −4x² + 12x = 4x(3 − x). Changing the last term from 2(x + 1)³ to 2(x − 1)³ changes the answer completely, from 8x² + 12x + 4.
The result is 2|xy| sqrt(5z) / z^3. Split off the even powers, keep absolute values on the odd-rooted variables, then rationalise the leftover sqrt(z) in the denominator.
The simplified form is 9|x| times sqrt(5). Factor 405 as 81 times 5, split the radical, and keep absolute-value bars because sqrt(x^2) equals |x|.
The simplified form is 2i sqrt(35). Pull out the imaginary unit first to get i sqrt(140), then split 140 as 4 times 35 and take the root of 4 outside.
The result is -2a^2 + 7a/10. The a^2 terms combine to give 1 - 3 = -2, and the a terms need a common denominator first: 3/2 - 4/5 = 15/10 - 8/10 = 7/10.
The expression is the constant 64. Setting a = 3x^2+4 and b = 3x^2-4 turns it into (a-b)^2, and a-b = 8, so every power of x disappears from the answer.
The simplified form is 0.055x − 3010. Expanding gives 78000 × 0.055 = 4290, and 1280 − 4290 leaves the constant −3010.
The answer is 2x^2 + 3x + 17. The two cubic terms cancel and the difference of squares contributes -x^2 + 16, so only a quadratic survives at the end.
Answer: 1.76566975(x + 1). Group the repeated factors as 1.15^2 and 1.3^2, multiply the constants into a single multiplier, then distribute if a linear form is wanted.
The whole expression reduces to 36x. Expand the two cubes so the x² terms cancel in pairs, handle the −3x product separately, and watch every x³ term disappear.
The result is x^2 - 6. Expanding the product gives x^2 - x - 6, and the loose +x cancels the -x exactly, leaving a binomial with no linear term at all.
The whole expression collapses to the constant 25. Recognise A² + B² − 2AB = (A − B)², subtract the two brackets, and see why every power of x cancels out.
Answer: 0 for every x. Expanding (x+4)^2 gives exactly x^2 + 8x + 16, so the bracket subtracts the same trinomial term by term and everything cancels.
The expanded form is x² − y² + 4y − 4, which regroups as x² − (y−2)² and factors to (x−y+2)(x+y−2) — a second difference of squares hiding inside.
The result is x^2 - 2y. The first product is the difference of squares x^2 - y^2, and the -y^2 cancels against the +y^2 coming from expanding y(y - 2).
The result is x^2/4 - 3x sqrt(5) - 15. Squaring the bracket gives x^2/4 + x sqrt(5) + 5, and tripling then subtracting leaves only a quarter of the x^2 term.
It simplifies to x - 3, valid for x not equal to -3. Factoring x^2 - 9 as (x - 3)(x + 3) exposes the common factor, but the original is undefined at x = -3.
Distributing -3 gives -3x^2 - 3xy + 6y, so the three families combine to -2x^2, -4xy and +7y. Watch the -3 x -2y = +6y term.
The answer is x^2 + 10. The power 3^2 is a number, not a variable term, so it becomes 9 and then merges with the 1 while x^2 stays separate.
The result is (x+5)/(x+2) with x ≠ 2 and x ≠ −2. Factor both trinomials, cancel the shared (x−2), and keep the restriction the cancelled factor leaves behind.
The result is x^(-ab), equivalently 1/x^(ab). The power-of-a-power rule multiplies the exponents, and the negative sign moves the whole thing into a denominator.
The slope is m = -8. Subtract the y-values to get -8, subtract the x-values to get 1, and divide; the steep negative value says the line falls 8 units per step right.
The slope is m = 2. Substituting into (y2 - y1)/(x2 - x1) gives (4 - (-4))/(6 - 2) = 8/4, where the double negative in the numerator becomes addition.
The slope is m = 1. The formula gives (7 - 5)/(10 - 8) = 2/2 = 1, so the rise and the run are equal and the line makes a 45 degree angle with the x-axis.
The answer is m = −1/3. The rise is +1 and the run is −3, so the negative sign comes from the denominator alone, not from either of the negative coordinates.
The answer is 17. Turn 2 log5(root x + 2) into log5 of a square, drop the logs since base 5 > 1, and the x terms cancel leaving 28 < 7 root x, so x > 16.
The integers are 31, 32 and 33, so the smallest is 31. Set the first to x, solve 3x + 3 = 96 to get x = 31 - which is already the answer the question wants.
The only solution is (4, 0). Substituting u = x - 2 gives 2/u + 3 = 2|u|; the positive case yields u = 2, and the negative case has no real root.
The roots -1/2 and 1/3 give p = 1/6 and q = -1/6. Substituting turns the second inequality into x^2 - x - 6 < 0, whose solution set is (-2, 3).
x = 87870/0.30 = 292900 exactly. The identical terms combine to 0.30x, and because 878700 is divisible by 3 the answer is a whole number with no rounding.
x = 963. Writing 0.6 as 27/45 and 5/9 as 25/45 shows the two x coefficients differ by only 2/45, so the constant gap of 42.8 is magnified 22.5 times.
x = 52500. Dividing by 0.8 gives 1 + x/10000 = 6.25, subtracting 1 leaves x/10000 = 5.25, and multiplying by 10000 finishes it. Check: 0.8 * 6.25 = 5 exactly.
x = 87870/1.7 = 878700/17 = 51688.24 rounded to two decimals. The two identical terms add up to 1.7x, and the exact quotient 51688 + 4/17 never terminates.
Answer: x = 2,505,589.39. Halve the right side to 1,625,000, multiply 0.85, 1.09 and 0.7 into the single factor 0.64855, then divide to recover x.
Answer: x = 2,063,426.56. Reduce the right side to 1,625,000, collapse the three multipliers into 0.787525, then divide - a two-discount, one-markup chain.
The answer is x = 2500/39, about 64.103. Dividing by 0.78 is the same as multiplying by 50/39, which keeps the answer exact instead of a rounded decimal.
The answer is x = 100. Dividing by 0.8 first gives x + 100 = 200 in one step, which is faster and cleaner than distributing the 0.8 across the bracket.
The solution is x = 325. Add 100 to both sides to get 0.8x = 260, then divide both sides by 0.8 — or equivalently multiply by 5/4, since 0.8 is the fraction 4/5.
The answer is X = 1900. Add 1200 first to get 0.8X = 1520, then divide - and dividing by 0.8 is the same as multiplying by 1.25, so the value grows.
The solution is x = 800. Adding 400 gives 0.8x = 640, and since 0.8 equals 4/5, multiplying 640 by 5/4 produces the answer with no decimal division.
The answer is x = 2254. The two x terms differ by only 0.01x, so the constant 22.54 is divided by 0.01 — a one-percent gap magnifies the root a hundredfold.
Answer: C = 1.35 + 0.15B and B = (C - 1.35)/0.15. Add 0.15B to isolate C; to get B instead, move C across, divide by -0.15, and watch the sign.
Answer: X = 225. Expand 1.8(X + 100) into 1.8X + 180, combine like terms to reach 0.8X - 20 = 160, add 20 to both sides, then divide by 0.8 to finish.
The solution is x > -2. Multiplying by 6 turns the inequality into 6 - 3(x + 6) < 2(2x + 1), which simplifies to -14 < 7x after collecting terms.
x = -175.86. Because 2.7421 is larger than 2.7, the reciprocal difference is negative, so 1/x = -0.0421/7.40367 and x itself must come out negative.
x = 16.2196, not 16.2214. Isolating 1/x gives (3.3 - 2.7421)/(2.7421 x 3.3) = 0.5579/9.04893, and inverting that single quotient finishes the problem.
x = 17.7272. Isolating 1/x gives (3.3 - 2.7821)/(2.7821 x 3.3) = 0.5179/9.18093, and taking the reciprocal of that fraction gives the answer directly.
x = 1. When two unit fractions are equal and neither denominator is zero, the denominators must match, so 3x + 11 = 9x + 5 and both sides become 14.
The answer is x = −1.3. Cross-multiplying gives 1 = 30x + 40, so 30x = −39; the root avoids the excluded value x = −4/3.
The solution is y = x/(8x - 1), with x not 0 and x not 1/8. Combining the fractions gives x + y = 8xy, and collecting the y terms lets y be factored out.
The solution is x = 1, y = −1/2. Setting a = 1/x and b = 1/y turns a non-linear system into an ordinary linear one that elimination solves in two lines.
Both denominators reduce to x^2 + 4x plus a constant, so v = x^2 + 4x gives v^2 + 8v + 7 = 0. Only v = -1 yields real x, namely -2 ± sqrt 3.
Since x(x+2) = (x+1)^2 - 1, setting u = (x+1)^2 turns the equation into 1/(u-1) - 1/u = 1/12, i.e. u^2 - u - 12 = 0. Only u = 4 is admissible, giving x = 1 or x = -3.
x > 2, that is the interval (2, infinity). Multiplying by the positive 3 gives 1 + x < 3x - 3, and collecting terms leaves 4 < 2x with the direction unchanged.
The real solutions are x = 0 and x = 0.0156549. Spotting the visible root x = 0 and locating the single minimum at x = 0.0079085 proves there are exactly two.
The solution is x = 0.9. Multiplying by 1 changes nothing, so the equation is simply 1 - x = 0.1, which is linear and has exactly one solution.
The solution is x < 5. Moving the x terms to the right keeps the coefficient positive, so the final division by 2 leaves the inequality direction untouched.
Both denominators expand to x^2 + 6x plus a constant, so u = x^2 + 6x turns the equation into 12/(u+5) + 15/(u+8) = 2, i.e. 2u^2 - u - 91 = 0 with u = 7 or -13/2.
x = 104022.99. Distributing the 13% bracket first turns the equation into 139732000 = 1343.28x, a single division once every x-term is collected on one side.
x = 103254.32. With only 2.72x loose on the left, the collected coefficient is 1353.28, and one division finishes the problem.
Answer: x = 12,110.83. Evaluate 12812 x 1.0398 = 13,321.9176, clear the denominator to get 13,321.9176 = 1.1x, then divide by 1.1 to strip the 10% margin.
The roots are x = 0, x = 1 and x = −4/3. Combine the two cube terms first, pull out the common 4x, then factor the leftover quadratic 3x² + x − 4.
The solution is x = 13. Add the numerator to get 169, multiply both sides by x to reach 13x = 169, then divide by 13 since 169 = 13 squared.
The answer is x = 96/7, about 13.714. Adding gives 192/x = 14, and 192/14 then has to be reduced by the common factor 2 to reach lowest terms.
The answer is t = 8. Distributing −4 over (t − 5) gives +20, not −20 — that sign is where most attempts at this equation go wrong.
The answer is x = 6960. Add the numerator first to get 348/x = 0.05, then cross-multiply - dividing by 0.05 is the same as multiplying by 20.
Every real number is a solution. After distributing the minus sign both sides reduce to 8x + 2, so the equation is an identity with infinitely many solutions.
The answer is x = 317/7.8 = 1585/39, about 40.64. Add the numerator first, then cross-multiply - and note the decimal answer never terminates.
Answer: x = 3200/3, about 1066.67. Rewrite 1.5 as 3/2 and multiply 1600 by the reciprocal 2/3; the decimal repeats, so the fraction is the exact answer.
Answer: x = 202.31 (exactly 176.0096/0.87). Simplify the constant to 176.0096, collect the x-terms into 0.87x, then divide to recover the gross total.
Answer: x = 202,309,885.06. Evaluate the constant as 176,009,600, subtract 0.13x from both sides to get 0.87x, then divide by 0.87 to gross the base up.
x = 16939/1.923 = 8808.63. Multiplying by x and adding x turns the equation into 16939 = 1.923x, which is the rule x = total / (1 + ratio) in disguise.
The answer is [-4, -3) union (1, 2]. The outer inequality gives two rays and the inner one a band, and the intersection is two short intervals.
x = -52/5, or -10.4. Multiplying by x + 10 gives 2 = -5x - 50, and the root is checked against the excluded value x = -10 before being accepted.
Answer: k = mg tan(a) / (4 pi^2 (R sin a - a)). Factor 2pi k out of the left side, then divide, so the two factors of 2pi combine into 4 pi^2.
The solution is x = -3. The left side is pure arithmetic: 5 - 8 = -3 and 2 + (-3) = -1, so -1 = 2 + x, and subtracting 2 from both sides gives x = -3.
Here x^2 + 1/x^2 = t^2 + 2 (note the PLUS 2), so the equation becomes 2t^2 - 11t + 12 = 0 with t = 4 or 3/2, giving the four roots 2 ± sqrt 5, 2 and -1/2.
The answer is x is about 1.71562. Mixing an exponential with a linear term admits no closed form, so bracket the root between 1 and 2 and bisect to converge.
The solution is x ≤ 0 or x ≥ 2 with x ≠ 3. Substituting t = 2ˣ turns the second denominator into the perfect square (t − 8)², and the inequality into (u + 4)(u + 7) ≥ 0.
The answer is x = 10. Expanding gives 320 - 20x = 240 - 12x, so 80 = 8x; removing 10 from each bracket makes the two weighted amounts balance at 120.
The answer is x = 5/7. Removing the bracket turns -(4x + 2) into -4x - 2, giving 16x - 4 = 6 + 2x, so 14x = 10 and the fraction 10/14 reduces to 5/7.
The answer is x = 10/3, about 3.33. Combining gives 30x = 100, and 100/30 must be reduced by the common factor 10 rather than left as it is.
The solution is x = 121/2, exactly 60.5. Combine like terms to get 30x = 1815, then cancel the common factor 15 from the numerator and denominator.
The solution is x = 200/3, about 66.67. Combining like terms gives 30x = 2000, and 2000/30 reduces to 200/3, which has no exact decimal form.
The solutions are x = (-34 + sqrt 2941)/21 and x = (-34 - sqrt 2941)/21. No integer pair multiplies to -1785 and adds to 68, so the quadratic formula is required.
Answer: X = -10.26. Simplify the coefficient to 500/27, subtract 2210 to get 190 = -(500/27)X, and the negative sign survives because 2400 exceeds 2210.
The answer is x = 16. Add the numerator to get 400/x = 25, then cross-multiply - the answer counts how many whole parts of size 25 fit into 400.
The answer is t = 20. Subtracting 100t leaves 150t = 3000, so the faster rate needs 20 units of time to overcome a fixed head start of 3000.
Answer: x = 5120/3, about 1706.67. Since 1.5 = 3/2, dividing by 1.5 is the same as multiplying by 2/3, which keeps the answer as an exact repeating decimal.
The solutions are x = (-1 + sqrt 7)/5 and x = (-1 - sqrt 7)/5. The discriminant is 700, and pulling out sqrt(100) turns 10 sqrt 7 over 50 into the reduced form.
Real answers: x = plus or minus sqrt((sqrt5 - 1)/5), about 0.497. The u = x^2 substitution gives u = (-1 +/- sqrt5)/5, and the negative root yields two imaginary solutions.
Answer: x = 2/3, a triple root. The left side is exactly (3x - 2)^3, since 27x^3 = (3x)^3 and 8 = 2^3 with the middle terms matching -3a^2b and +3ab^2.
Answer: x = 216.809. The fraction 0.05/0.95 reduces to exactly 1/19, so the equation collapses to 228.22 = (20/19)x, giving x = 228.22 times 19/20.
x = 298/13 = 22.9231. Collecting terms gives 1.3x = 29.8, and clearing the decimals shows the exact value is 298/13, which is a non-terminating decimal.
The answer is x ≈ 281.46. Collecting the x terms gives 1.01543461x = 285.8, because the fee fraction 0.0152/0.9848 is just another coefficient of x.
Answer: x = 294.9308, not the 111.241 sometimes quoted. The two 1.52% factors multiply, so the coefficient is only 0.00023461 and x barely moves from 295.
x = 2981/5.3043 = 562.00. Adding x to both sides gives 2981 = 5.3043x, so x = total / (1 + ratio); the leftover 2419.00 is then 4.3043 times x exactly.
The answers are x = 1 and x = 2. A line can cross a convex exponential curve at most twice, so once both integer roots are found the search is provably complete.
The roots are x = 1 and x = 1/3. Recognise (2x − 1)² − x² as the difference of squares a² − b², factor into (x − 1)(3x − 1), then use the zero-product property.
x = 1/2 or x = -7/3. The equation is already factored, so set each bracket to zero separately and solve the two linear equations that result.
The solution is x <= 4, or (-inf, 4]. Add 3 to undo the subtraction, then divide by the positive 2, which leaves the inequality direction unchanged.
The solutions are x = 2 and x = 4/3. Both terms are already squares, so factoring as (A − B)(A + B) gives the two linear factors x − 2 and 3x − 4 directly.
The answer is x < 51/28, about 1.8214. Multiplying every term by the LCM 12 removes both fractions at once, turning it into 24x - 15 < 36 - 4x.
The solution is x > −1/2. Multiplying by the LCM 12 removes both fractions at once, and since 12 is positive the inequality sign never flips.
The solution is x > -4. After expanding, the x terms collect on the right to give -4 < x, which is then simply reread with the variable written first.
The solution is x < 7/10. Collect the x terms on the left and the fractions on the right, then add 1/2 and 1/5 over their common denominator of 10.
The solution is x <= 0. The bracket is preceded by a minus sign, so both inner terms change sign, giving 2x - x + 1, which collapses to x + 1 <= 1.
The answer is x > 5. Subtract x to leave x + 1 > 6, then subtract 1; the boundary x = 5 makes both sides equal 11, so it is correctly excluded.
The answer is x > 6. Subtract x from both sides to leave x + 1 > 7, then subtract 1; because no negative divisor appears, the sign never flips.
The answer is x = 72.3. Evaluate 13×18 = 234 and 6×4.4 = 26.4 before touching x, so the equation reduces to 2x + 260.4 = 405 in one step.
The answer is x = 4. Subtract 2 from both sides to get 2x = 8, then divide by 2; undoing the addition before the multiplication is what keeps it simple.
The solution is x = -mu / (2(1 + lambda)) when lambda is not -1. At lambda = -1 the x terms cancel, leaving no solution unless mu is 0, when every x works.
The solution is x = −μ/(2(1 + λ)) whenever λ ≠ −1. If λ = −1 the x terms vanish: every real x works when μ = 0, and there is no solution when μ ≠ 0.
x = 2. Expanding (2x + 3)^2 makes the 4x^2 terms cancel, so the quadratic-looking equation collapses to the linear equation 12x + 9 = 51 - 9x.
The solution is x = 2, y = 1. Solve the second equation for x = y + 1, substitute into the first to get 5y + 2 = 7, then back-substitute to recover x.
The solution is x = 4. Subtracting 5 from both sides gives 2x = 8, dividing by 2 gives x = 4, and substituting back returns 13 exactly as required.
The answers are x = (1 + root 3)/2 and (1 - root 3)/2. The discriminant 12 is not a perfect square, so the roots are irrational and factoring cannot work.
The solutions are x = 1 and x = 3/2. With a leading coefficient of 2, split -5x using factors of 2 times 3 = 6 that sum to -5, namely -2 and -3.
The roots are x = 3 and x = -5/2. Find two numbers with product ac = -30 and sum -1 (namely 5 and -6), split the middle term, and factor by grouping.
The roots are x = 2 and x = -3/2. Find two numbers with product ac = -12 and sum -1, namely 3 and -4, split -x into 3x - 4x, then factor by grouping into (2x+3)(x-2).
The solution is x < -1 or x > 1/2. Factoring 2x^2 + x - 1 as (2x - 1)(x + 1) exposes the two roots, and an upward parabola is positive outside them.
The answer is y <= (40 - 3.8x)/7.7, roughly y <= 5.19 - 0.49x. Dividing by the positive 7.7 keeps the direction, so the region sits below the boundary line.
The answer is x > log base 4 of 3 x 10^9, about 15.74, so the least integer is 16. Since 4^x is always positive, no inequality sign ever flips.
The answer is x = -10/7, or about -1.4286. Distributing the -2 over (2 - 5x) gives -4 + 10x, so the equation becomes 3x - 9 = 1 + 10x, hence -10 = 7x.
The answers are x = 27 exactly and x is about 1.15082. Taking logs reduces it to ln(x)/x = ln(3)/9, a curve that meets a horizontal line twice.
The solution is x < 0.0000020000004 or x > 14.3704368. The convex function 3^x - 500001x dips below zero once, so the line is crossed at exactly two places.
The answer is x = 5 + 3^(49/32), about 10.3776. Isolate the log first, convert to exponential form, then confirm x > 5 so the logarithm is defined.
The answer is x = 5.455/32.63, about 0.16718. Subtract the constant first, then divide - two operations, and the decimals never need clearing.
The answer is x = 10.455/32.63 = 0.32041, which rounds to 0.3204 - not 0.3203. The fifth decimal digit is a 1, so the fourth place rounds down.
The solution is X = 27. Cross-multiplying gives 3150 + 54X = 7200 - 96X, so 150X = 4050; both sides of the original equation then equal 32.
The solution is -1 < k < 1. Dividing by the positive 36 leaves 1 - k^2 > 0, and k^2 < 1 means k lies strictly within one unit of zero on the number line.
x = 1800 + 900 sqrt 7 = 4181.18 or x = 1800 - 900 sqrt 7 = -581.18. The discriminant 22680000 equals (1800 sqrt 7)^2, so both roots simplify neatly.
The solutions are a = -3 and a = -5. Move -45 across, divide the whole equation by the common factor 3, then factor a^2 + 8a + 15 as (a + 3)(a + 5).
The roots are x = 0, x = −2 (double) and x = 4/3. Grouping fails on this quartic; the rational root x = −2 and division give the true factorisation.
The single solution is x = 4/9. Both products contain 9x², so expanding makes the quadratic terms cancel and what looks like a quadratic is really a linear equation.
Answer: no integer solution; the real roots are x = 8 + or - (14)^(1/4)/3, about 8.6448 and 7.3552. Dividing by 7^3 leaves (3x - 24)^4 = 14, which is not a fourth power.
The solution is x >= 1. Adding the coefficients 3 and -2 leaves a single x, so the inequality reduces to x - 1 >= 0 and needs only one more step.
The solution is x > 1. Distribute the 2 over x - 2 to get 2x - 4, subtract 2x from both sides, and the single remaining x needs no division at all.
Answer: x = 3, a double root. With u = 3x - 5 the equation is u^2 - 8u + 16 = (u-4)^2 = 0, so u = 4 and 3x = 9.
The solutions are x = (5 + sqrt 43)/3 and x = (5 - sqrt 43)/3. Multiplying through by x turns the rational equation into the quadratic 3x^2 - 10x - 6 = 0.
The solution is x > 15. Subtracting 2x leaves a coefficient of exactly 1, so after moving the constant 15 across the answer is read off directly.
The answer is x < 4. Multiplying through by 4 gives 3x + 4 > 8x - 16, and collecting on the right avoids ever dividing by a negative number.
The answer is x = 10, and not 25/3. The right side is 35, so 3x = 30 - the equal constants on both sides cancel and never enter the division.
The answer is x = -7, the only solution. The right side can never be positive, so 3x + 5 <= 0 forces x <= -5/3 and only one case has to be checked.
The solution is x = 5. Subtract the 7 that was added last, leaving 3x = 15, then divide by the 3 that multiplied x first, and substitute back to check.
The answer is x < 2/3. Subtract 3x so the x coefficient stays positive, subtract 7, then divide by 3 — no sign flip is needed at any stage this way.
The answer is x < 3/4. Subtract 3x so the coefficient stays positive, subtract 6 to get 3 > 4x, then divide by 4 and read the inequality backwards.
The answers are x = 4 and x = -4. With no middle term you can isolate x^2 = 16 and take roots directly, but the plus-or-minus sign is mandatory.
x = 0 or x = 1/3. There is no constant term, so x is a common factor; dividing both sides by x instead would silently destroy the root x = 0.
The answers are x = (-5 + root 97)/6 and (-5 - root 97)/6. Since 97 is prime and not a perfect square, the roots are irrational and cannot be factored.
The solution is x < 0. Put every term over the common denominator 6, multiply through by 6, collect like terms to get -12x > 0, then divide by -12 and flip the inequality sign.
The roots are x ≈ −3.1660 and x ≈ 2.9426. Their sum −0.2234 equals −b/a and their product −9.3160 equals c/a, which validates both decimal answers at once.
The roots are x ≈ −2.6457 and x ≈ 2.5303. Vertex form is −4.4912(x + 0.05769)² + 30.0809, so x + 0.05769 = ±√(30.0809/4.4912) = ±2.5880 gives both roots.
The roots are x ≈ −2.1144 and x ≈ 2.0515. Multiplying through by −1 gives 4.6173x² + 0.2906x − 20.029 = 0, which keeps every sign in the formula positive.
The solution is m < 7/3. Group the constants 4 and 3 into 7 and the variable terms -2m and -m into -3m, then move -3m across to avoid a sign flip.
The answer is x = 1123/390 ≈ 2.879487. Subtract 0.68175, then divide by 4.9725 — converting to a fraction avoids the rounding slip to 2.8791.
The answer is x = 77273/19890 ≈ 3.885018. Subtract 0.68175 to get 19.31825, then divide by 4.9725 and keep the exact fraction before rounding.
The solution is x > −7. On the right the +6x and −6x cancel, leaving the constant −12; dividing by −4 at the end reverses the inequality sign.
The solution is x >= -6. Expanding both sides gives 4x + 4 >= 3x - 2; because you only ever add or subtract, the inequality sign never flips.
The solution is x = log base 4 of 80, about 3.1610. Since 80 is not a power of 4, taking logarithms and dividing ln 80 by ln 4 gives the exact and decimal answers.
The roots are x = −25 ± 5√34, about 4.154759 and −54.154759. Collect 100x, halve the equation to x² + 50x − 225 = 0, then use the quadratic formula.
x = 3/10 or x = -11/4. The discriminant 14884 equals 122^2, a perfect square, so the roots are rational and 40x^2 + 98x - 33 factors as (10x - 3)(4x + 11).
The answer is x = 125/7, about 17.857. Dividing gives 750/42, and the two share a factor of 6, so the fraction reduces exactly once and then stops.
The roots are x = 2.43 and x = 1.23. Move everything to one side to get 3x^2 - 11x + 9 = 0, then apply the quadratic formula with discriminant 121 - 108 = 13.
x = 4402/73, about 60.3014. Distributing the minus sign gives 3.65x + 0.35 = 220.45, so 3.65x = 220.10, and the quotient is not the round number 60.3.
x = 4405/73, about 60.3425. After distributing, the equation reads 3.65x + 0.35 = 220.6, so 3.65x = 220.25 and dividing gives a non-terminating decimal.
x = 4910/73, about 67.2603. The bracket expands to -0.35x + 0.35, leaving 3.65x = 245.50, and 73 being prime keeps the answer a non-terminating decimal.
The answer is x = 126618/1825 ≈ 69.379726. Distributing −0.35 gives +0.35, so the equation reduces to 3.65x + 0.35 = 253.586.
The answer is x = 55617/730 ≈ 76.187671. After distributing, the equation becomes 3.65x + 0.35 = 278.435, so x = 278.085/3.65.
x = 8. Distributing gives -60 + 3x, so the equation becomes 7x - 60 = -4; adding 60 and then dividing by 7 leaves x = 8, and 32 - 36 = -4 confirms it.
The answer is x = -8. Subtract 4x to gather the variable on the right, subtract 9, then divide by 2; both sides then evaluate to -39 as a check.
The solution is x < 2. Adding 2x to both sides gives 6x + 8 < 20, then 6x < 12, so dividing by the positive 6 keeps the inequality direction.
The solution is every real number except x = 1/2. The left side factors as the perfect square (2x - 1)^2, which vanishes only at its repeated root x = 1/2.
The solutions are x = (1 + sqrt 33)/8 and x = (1 - sqrt 33)/8. Since 33 = 3 times 11 is squarefree, the radical cannot be simplified and the fraction stays as is.
The discriminant is 256 - 112 = 144, a perfect square, so x = (-16 +/- 12)/8 gives -1/2 and -7/2 - and the quadratic also factors as (2x+1)(2x+7).
The two solutions are x = -3/2 and x = -7/2. The discriminant is 400 - 336 = 64, a perfect square, so both roots come out as exact fractions.
The solution is x = 12. Inside the bracket 13x/4 - 11x/6 combines to 17x/12, and that 17 cancels the divisor, leaving the equation 5.6 - x/4 = 2.6.
The solution is x = 3. Distributing -2 across x - 1 gives -2x + 2, so the equation becomes 7 - 2x = 1, then -2x = -6, and finally x = -6 / -2 = 3.
The solution is x <= -2. Distributing gives 3 - x/2 >= 4, hence -x/2 >= 1, and multiplying both sides by -2 reverses it, producing x <= -2 not x >= -2.
The answer is n = 12/5 = 2.4. Distributing the leading minus turns −(−5n + 1) into +5n − 1, so the two n terms add rather than cancel.
The root is x = -31/5 = -6.2. Dividing both sides by 5 first turns the equation into x + 8 = 9/5 and avoids expanding to 5x + 40 = 9 altogether.
The answer is x = (5ln3 - 2ln5)/ln15, about 0.8398 - and not 1.037. Different bases force logarithms, and ln5 + ln3 collapses neatly into a single ln15.
The answer is x = −2934271/5900 ≈ −497.334068. Writing 1 as 113/113 merges 5x/113 and x into 118x/113, and the large constant forces a large negative root.
x = 5399/5.8116 = 929.00. Clearing the fraction gives 5399 = 5.8116x, so x = total / (1 + ratio); the leftover 4470.00 is then exactly 4.8116 times x.
The solution is x = 5. There is no constant term to move, so dividing both sides by 580 finishes the problem, and 2900/580 reduces exactly to 5.
The solution is a < -2 or a > 6/5. The discriminant 256 is a perfect square, giving the roots -2 and 6/5, and the upward parabola is positive outside them.
The answer is w = -7z/2. Collect the w terms on the left and the z terms on the right to reach 2w = -7z, then divide by 2 without touching the sign.
The solution is x = 5. Subtracting 2x leaves 3x - 4 = 11, adding 4 gives 3x = 15, and dividing by 3 yields x = 5, which both sides confirm as 21.
The solutions are x = −2 and x = 1/3. Recognising 49x² as (7x)² turns the equation into A² − B² = 0, which factors into two linear equations instead of a quadratic.
The answer is x = 0.8. Expanding cancels 25x² on both sides, so the apparent quadratic collapses to the linear equation −80x + 64 = 0.
The exact solution is x = -1. Subtract 7 from both sides to reach 5x = -5, then divide by 5; substituting back gives 5(-1) + 7 = 2, so the answer checks out.
The solution is x > -5/4. Combining 5x and -2x gives 3x, and dividing the fraction -15/4 by 3 reduces to -5/4 with the inequality direction preserved.
The solutions are x = 2 and x = -2. With no linear term you can isolate x^2 = 4 and take roots directly, which is faster than the quadratic formula here.
The answer is x = 280 exactly. The rate gap is only 0.05, and 14/0.05 is a whole number because dividing by 0.05 is the same as multiplying by 20.
The answer is x = 14/0.17, about 82.35. The two x terms nearly cancel, leaving only 0.17x - which is why a small constant produces a large answer.
The answer is p = -7. Expanding both sides gives 30 + 6p = 7p + 37, and subtracting 6p leaves p on its own with no division step needed at all.
The answer is x = −23. Expanding gives 6x − 30 = 7x − 7; moving the x terms right and the constants left leaves x = −23 in a single step.
The three real roots are about -0.91311, about 1.11768, and exactly 36, because 36^18 = (6^2)^18 = 6^36. The function ln(x)/x controls the two positive roots.
Answer: X = 4347/50 = 86.94. Combine the two fractions into 500/27, subtract 2210, and this time the left-hand side is smaller so X comes out positive.
The answer is g = 18. The three g terms collapse to 2g because the bare g counts as 1g, so the equation becomes 2g = 36 before any division.
The solution is p < (3 - sqrt 21)/6, or 0 < p < 1/2, or 1 < p < (3 + sqrt 21)/6, because the quintic factors neatly as p(p - 1)(2p - 1)(3p^2 - 3p - 1).
The roots are x = (-5 + sqrt 61)/4 and x = (-5 - sqrt 61)/4. Expand the bracket to reach 4x^2 + 10x - 9 = 0, then divide the quadratic formula through by 2.
Answers: s = -6 and s = -64/7. Substituting u = s + 9 turns the equation into 7u^2 - 20u - 3 = 0, which factors as (7u+1)(u-3) = 0.
The answer is k = 34.5. Subtracting 300 first gives 200k = 6900, and since 6900 is not a whole multiple of 200 the answer lands on an exact half.
The discriminant of x^2 - 37x + 345 is -11, so there is no real answer. The maximum of 74x - 2x^2 is 684.5, which falls just short of the 690 target.
Subtract 4 and combine into (7 - 5x)/(3x - 2) < 0, then use a sign chart. The solution is x < 2/3 or x > 7/5; multiplying both sides by 3x - 2 would be invalid because its sign changes.
The answer is x >= 7. Subtract 4x, add 15, then divide by the positive 3 - because the divisor is positive, the inequality sign never has to flip.
The solution is x > -2. Distribute the 3 across the bracket, collect the x terms on the left and the constants on the right, then divide by the positive 4.
The solution is x = -5. Subtracting 3x collects the variable on the left as 4x, then moving the 8 across gives 4x = -20 and dividing by 4 finishes it.
The answer is x = −7.6. Multiply both sides by (x + 6), expand −5(x + 6) = −5x − 30, and check that the root does not make the denominator zero.
The answer is x = 2.8/1020000 ≈ 2.745098 × 10⁻⁶. Huge coefficients on both sides collapse to 1020000x, so the root is six orders of magnitude below 1.
Answer: x = -0.0058737255. Adding 10000x to both sides gathers the coefficient into 1020000, and the constant gap of -5991.2 makes x small and negative.
Answer: x = 7300/13, about 561.54. Distribute both brackets, collect 29200 - 27x, move the x-terms to one side to get 52x = 29200, then divide.
The solution is x = 1.55095. Because the same number multiplies x and stands alone, factoring gives x + 1 = 215900.9/84635.579, one division instead of two steps.
g > 5. Distributing gives 2g - 6 < 8g - 36, and moving the g terms to the right keeps the coefficient positive, so 30 < 6g with no sign flip needed.
The answers are x = 0 and x = 2. Move everything to one side and factor out 8x; dividing both sides by x instead would silently destroy the root x = 0.
The answer is a - b = -4. The individual values of a and b cannot be found from one equation, but their difference is pinned down by subtracting 9 from both sides.
The roots are x = 5/3 and x = 1/3. The equation is already a perfect square, so isolate (x-1)^2 = 4/9 and take both square roots rather than expanding into a quadratic.
x = log base 9 of 252 = ln 252 / ln 9 = 2.5166. Multiplying by 12 isolates 9^x = 252, which is not a power of 9, so a logarithm is unavoidable here.
x = 9127/1.934 = 4719.23. Multiplying by x and then adding x gives 9127 = 1.934x, which is the rule x = total/(1 + ratio) with a ratio just under 1.
Answer: x = 7517/7, about 1073.86. Divide both sides by 100, isolate the fraction to get 0.08 = 3/(0.14(x - 806)), then invert it and add 806 back on.
x ≈ 1.2503. Both bracketed products share the factor 4 − 2x, whose combined coefficient is 41.4 + 28 = 69.4, reducing everything to 305.6 − 122.8x = 152.0621.
Answer: Vout = 4.81748. The linear coefficients 96.844679 and 97.109892 almost cancel, so every digit matters; exact fractions give -0.265213 V - 3.6354e-7 V^3 + 1.2777 = 0.
The roots are -0.01062039, 0.02187147 and 245.69974892. Two roots are near zero and one is huge, so Vieta checks alone cannot confirm the small ones.
Answer: y > 3/8. Clear the decimal denominators 0.03, 0.5 and 0.15 by rewriting each quotient as a fraction, combine like terms, then divide by 8 to get y > 3/8.
The real root is x = 0.0536118. Bracketing must use correctly evaluated test points: f(0.06) is +0.0246, so the sign change lies below 0.06, not above it.
Split every fraction into a coefficient and a constant, collect to 18.3451x = -13.2857, and divide to get x = -0.7242, i.e. -0.72 to two places.
Answer: X = 14487500/1313781, about 11.027. Clear the products 3.6 x 7 x 60 = 1512 and 25.2, collect the X terms, then divide 11.498016 by 1.0426833.
Expand and split each fraction into a decimal coefficient plus a constant, collect to 20.7775x = 49.9345, and divide to get x = 2.4033, i.e. 2.40 to two places.
The roots are x = 15.97929 and x = 8.02999. The discriminant is 0.5945668496; a single mistyped digit in 4ac shifts both roots in the third decimal place.
The roots are x = 1 (double), -1, -5 and plus or minus i sqrt 2. A rational root, then grouping, then a substitution y = x^2 break the sextic apart.
Clearing decimals gives y^3 + 26y^2 + 25y - 100 = 0 with y = x^2. Its one positive root is y = 1.50569, so x = +/- 1.22706 are the only real solutions.
The answer is x is about 5.666. Peel the logarithms one at a time by exponentiating base 10 twice, after collapsing the right-hand side to a single number.
The roots are x ≈ −1.4580 and x ≈ 1.4395. The discriminant is 191.131, and its square root 13.8250 dominates the tiny linear coefficient, so the roots are nearly symmetric.
The solution is x = -1309/110 = -11.9. Evaluate the four triple products first, combine the constants to 1309, then divide by the coefficient 110.
x = 193.92W/0.87 = 222.90W. The unknown appears on both sides, so the 0.13x must be collected before dividing - it is a linear equation, not an evaluation.
The solution is q = 5. Combining the two constants on the left into -20 first, then subtracting 4q from both sides, turns the equation into 6q = 30.
Answer: x = 591.4491. Because the same coefficient appears twice you can factor it out, giving x + 1 = 215900.9/364.4210 and a one-line answer of x = 591.4491179.
The solution is x = 6710/0.97 = 6917.53. Collapsing the bracket to x + 1000 first turns the whole equation into 0.97x - 210 = 6500 in a single step.
x = 2. Expanding all four brackets gives 200 - 56x = 18x + 52, so 148 = 74x. Both sides of the original equation equal 88 at x = 2.
The solution is x = 12/5 and y = root 6 over 5. Scaling the equations by root 2 and root 3 turns both y-coefficients into root 6 so they cancel on addition.
The answer is x = plus or minus root a when a > 0. Multiplying by x gives x^2 = a, but x = 0 is barred by the original fraction, so a = 0 has no solution.
x = (2880 - m)/9 and y = (1440 - 2m)/9, and the inequality x + y <= 8m/3 holds exactly when m >= 160. One sign slip when isolating x reverses the whole condition.
The solution is x = 2201834.869. Turning 3% and 13% into 0.03 and 0.13 makes the x terms combine to +0.10x, since -0.03x + 0.13x leaves a tenth.
The solution is x <= 47.94, or x <= 47 for whole numbers. Dividing by the negative coefficient -0.85 is what flips the inequality sign.
Answer: x = 2000.8486. Evaluate 0.35 x 1077.38 = 377.083 first, subtract it, then divide by 0.35; equivalently x = 1077.38 x 0.65/0.35 in a single line.
The solution is a = 12. Adding -16 is the same as subtracting 16, so the equation becomes a - 16 = -4, and adding 16 to both sides isolates a.
Expanding gives the cubic −0.013122x³ + 50.058x² − 41762.7x + 6482700 = 0, whose three real roots are x = 24500/27 ≈ 907.4074, x ≈ 201.1822 and x ≈ 2706.2252.
The solution is x = -385.8956. Combine the numerator to 1202.61 - x, cross-multiply to 9(1202.61 - x) = 16(1279.43 + x), then collect the x terms.
The solution is x >= 1/2. Expanding both sides gives x^2 + 4x + 3 >= x^2 - 6x + 8, and subtracting x^2 leaves an ordinary linear inequality.
Factoring out z leaves the depressed cubic z^3 + 2r^2 z - 8 = 0 with r^2 = x^2 + y^2, whose single real root Cardano formula gives explicitly.
The solutions are x = 0.9 exactly and x = 0.8974433. Factoring (8.9-x)^2 - 64 as (0.9-x)(16.9-x) exposes a common factor and reduces a quartic to a cubic.
The answer is x ≈ 4.9368×10⁻⁵. Evaluate the known reciprocal cube as 8×10¹², add it across, then invert and take a cube root — all in scientific notation.
The solution is x = 24. Collect the three constants on the left into -4 and the two x terms on the right into -12x, then move terms to get 3x = 72.
The only real solution is s = -3.0864085. Clearing denominators gives 2s^3 + 33s^2 + 180s + 300 = 0, which has no rational root despite appearances.
The two real roots are x = 0.0221841 and x = -2.1184421. Factoring out (1+x)^27 turns the ten powers into a geometric series, then a sign change brackets each root.
The solution is a = −1/550, b = 199/1100, c = 61/55. Because c has coefficient 1 in all three equations, subtracting consecutive pairs removes it in one stroke.
Answer: x1 = 19, x2 = -8, x3 = 1. Swap rows to get a leading 1, clear below the pivots, then back-substitute to reduced row echelon form and read off the solution.
D = 12830.69 and W = 12169.31. Substituting W = 25000 - D collapses the second equation to 0.945D = 12125, since the two rates differ by 0.945.
The solution is X = 4963/11 ≈ 451.18 and Y = 12142/11 ≈ 1103.82. Multiplying the second equation by 5.15 and subtracting isolates Y in a single step.
The solution is X = 2393/18 ≈ 132.94 and Y = 3218/9 ≈ 357.56. Substituting X = 490.5 − Y turns the system into one equation in Y with coefficient 0.45.
The solution is X = 4481/11 ≈ 407.36 and Y = 13064/11 ≈ 1187.64. The coefficient determinant is 5.20 − 5.75 = −0.55, which drives both quotients.
The solution is X = 524.625 and Y = 1103.375. The average price is 6627.75/1628 ≈ 4.0711, and its position between 3.80 and 4.20 fixes the split directly.
The solution is x = -339.79. The x coefficients differ by only 0.14394, so the constant gap of -48.9091 is magnified about seven times when you divide.
The solution is x = -374.70. Subtracting the smaller x term leaves a coefficient of only 0.12987, so a small constant difference produces a large negative root.
The answer is exactly x = 37.8, not just an approximation: 52.5/76.25 reduces to 42/61 and 549 = 9 × 61, so 54.9 · 42/61 collapses to 378/10 with no rounding.
The answers are x = -6 and x = 10. Simplify the right side to 8 first, then split into 2 - x = 8 and 2 - x = -8 to catch both of the solutions.
The answer is the closed interval -7/2 <= a <= 1/2, not isolated points. Between the two kinks the a terms cancel and the equation becomes 8 = 8.
The answer is x = 3/4 only. The right side must be non-negative, so x >= 0, and that constraint eliminates the second case's candidate x = -3/8.
The solution is -5/2 < x < 4. The breakpoints -1/3 and 4 give three cases, and the last one collapses to x + 1 < x + 1, which is never true anywhere.
The solutions are x = 16/3 and x = −4/5. When two absolute values are equal the insides are equal or opposite, so split into 4x − 6 = x + 10 and 4x − 6 = −x − 10.
x = 4/7 or x = -2/7. Because 3 is positive the equation splits into 7x - 1 = 3 and 7x - 1 = -3, so the two roots land 3/7 on either side of x = 1/7 exactly.
The answer is -2 < x < 2. Between the kinks the sum is constantly 2, and outside it grows by 2 per unit, so it reaches 4 exactly one unit past each kink.
The answer is x < 0 or x > 7. Between the kinks the sum is constantly 3, so the total only reaches 7 two units beyond each of the two kink points.
The answer is x < -3 or x > 0. The middle interval collapses to the false statement 1 > 3, so the solution set is the two outer rays and nothing between.
The answer is x = 11. The equation says x is equidistant from -66 and 88, so it must be their midpoint - one line instead of two algebraic cases.
The only solution is x = 3/2. Breaking the line at 0, 1, 2 and 3 gives five cases, and four of them collapse to contradictions or roots outside their own interval.
The solution is a = 78.77. Taking 13% of 10% means multiplying by 0.013, so the deduction is only 0.091a and the a coefficient stays positive at 6.909.
The solutions are b = 4 + sqrt 11 and b = 4 - sqrt 11. Rename the formula's coefficients to A, B, C so the variable b is never confused with the middle coefficient.
The solution is x = 2034022.99. Moving 0.13x across leaves 0.87x = 1769600, because the right-hand x has an unwritten coefficient of exactly 1.
x = m, provided G and d are nonzero. Both sides carry the same denominator d^2 and the same factor G, so cancelling leaves Gm = Gx and therefore x = m.
The solution is x = (4k - 6)/(k - 3) for k not equal to 3. When k = 3 the equation reduces to 6 = 12, a contradiction, so there is no solution at all.
x = exp(5 ln2 ln3/(ln2 + ln3)) = 8.3729, not 6.06. Converting both logarithms to natural logs turns the equation into a single linear equation in ln x.
The exact solution is x = −22703/143 ≈ −158.7622. Subtract 0.06997 to get −0.00143x = 0.22703, then divide by the negative coefficient, which flips the sign of the answer.
The roots are x = (5 ± √785)/20, about 1.650893 and −1.150893. The discriminant 785 = 5 × 157 has no square factor, so the surd cannot be simplified.
x <= (65 - 5 sqrt 161)/2 = 0.7786 or x >= (65 + 5 sqrt 161)/2 = 64.2214. Dividing by -4 reverses the inequality to x^2 - 65x + 50 >= 0, true outside the roots.
x = -3. The bare leading minus turns -(5x - 4) into -5x + 4, and after collecting terms -3x - 10 = -1, so -3x = 9, and dividing by -3 then gives x = -3.
The answer is x < 7. Dividing by the negative -6 reverses the direction, so the greater-than becomes a less-than - the single step that decides the answer.
The answer is x = -2. Simplify the right side to 14 first, then divide by -7 - a positive divided by a negative gives a negative quotient, not 2.
t = -3 + sqrt(37), about 3.0828. Multiplying through by -1 gives t^2 + 6t - 28 = 0 with roots -3 +/- sqrt(37), and the condition t > 0 rules out -3 - sqrt(37).
There is no real solution; the complex roots are m = (1 + i sqrt 3)/2 and (1 - i sqrt 3)/2. Clearing m gives m^2 - m + 1 = 0 with discriminant -3.
The solution is x = 1/(2 - m), valid whenever m is not 2. Isolate the reciprocal, flip both sides, and note that m = 2 makes the denominator vanish.
The solution is 9/2 <= x < 5. Subtracting 8 from all three parts gives -10 < -2x <= -9, and then dividing by -2 reverses both inequality signs at once.
The answer is x = 3. Add 9 to both sides to get -2x = -6, then divide by -2; two negatives divide to a positive, so the answer is positive 3.
The answer is x <= -2 or x >= 2. Dividing by the negative -2 reverses the sign to x^2 >= 4, and a greater-than square gives two outward rays.
The solution is x >= -3. After subtracting 6 you divide by -4, and dividing an inequality by a negative number reverses its direction from <= to >=.
x = 2 or x = 3. Pulling out the common factor -4 leaves the simple monic quadratic x^2 - 5x + 6, which factors as (x - 2)(x - 3).
The roots are m = 3.67 and m = -0.27. Move every term to the side that makes the m^2 coefficient positive, giving 5m^2 - 17m - 5 = 0 with discriminant 389.
The answer is x < -4/3. Dividing both sides by -6 reverses the inequality, and the resulting fraction -8/6 reduces to -4/3, or about -1.333.
The solutions are t = -3 + sqrt 37 and t = -3 - sqrt 37. Multiplying by -1 gives t^2 + 6t - 28 = 0, whose discriminant 148 simplifies to 2 sqrt 37.
The answer is x < 1. Moving the 1 across gives -x > -1, and dividing by -1 flips the greater-than into a less-than, which is the whole difficulty here.
The solution is 1 <= x <= 3. Multiplying by -1 reverses the inequality into x^2 - 4x + 3 <= 0, which holds between the roots 1 and 3 inclusive.
There is no solution. Flipping to x^2 - 6x + 10 < 0 gives a discriminant of -4, so that upward parabola never reaches the axis and is always positive.
The roots are x = 3 and x = -1/2. Multiplying by the LCD of 30 gives 2x^2 = 5x + 3, and factoring 2x^2 - 5x - 3 as (2x + 1)(x - 3) finishes the job.
x < -4, that is the interval (-infinity, -4). Multiplying both sides by the positive number 3 keeps the inequality direction; only a negative multiplier flips it.
The four roots are x = −1, 3/2, 2 − √3 and 2 + √3. Factor the quartic with the rational root theorem and synthetic division, then use the quadratic formula.
The unique solution is x1 = 2, x2 = 2, x3 = 0. Eliminating x1 from the second and third equations leaves a 2x2 system that collapses immediately to x3 = 0.
The unique solution is x1 = 5, x2 = 3, x3 = -1. Two of the equations are missing a variable, so expressing x1 and x2 in terms of x3 solves the whole system fast.
Read the left side as the total distance from x to 0 and to 1. That sum equals the gap 1 exactly when x lies between them, so the solution is the whole interval [0,1].
The solutions are x = 9 and x = -9. The outer bars split into |x| - 4 = 5 and |x| - 4 = -5; the second gives |x| = -1, which no real number satisfies.
Substituting t = x^2 gives t^2 - 24t - 25 = 0 with t = 25 or t = -1. Only t = 25 yields real x, so x = 5 or x = -5.
Answer: x = 24.5. Expand to 367.5 + 4x = 372.4 + 3.8x; the x terms differ by only 0.2, so a gap of 4.9 in the constants needs 24.5 units of the new material.
The answer is -3 < x < -2. Subtracting 4 from all three parts is the only step needed, and both bounds land negative because 4 exceeds them both.
The roots are x = −3, x = 1/2 and x = 1. The rational root test flags x = 1/2, so 2x − 1 divides out and leaves x² + 2x − 3 = (x + 3)(x − 1).
The answer is x > cube root of 5, about 1.71. A cube root is defined for all reals and is strictly increasing, so the answer is one ray with no case split.
The answers are x = 1 (a double root) and x = -2. Synthetic division by x - 1 leaves x^2 + x - 2, which contains the factor x - 1 a second time.
The real root is x ≈ −0.6535952, with complex roots x ≈ −0.6732024 ± 2.3805031i. No rational root exists, so bracket between −1 and 0 and refine with Newton's method.
The real answer is x = 7, since the cubic is x(x+1)(x+2) = 504 and 7*8*9 = 504. The other two roots are the complex pair -5 plus or minus i root 47.
Answer: x = (-3 +/- sqrt(555))/2, about 10.279 and -13.279. Move 2.91 across, multiply by 100 to get 2x^2 + 6x - 273 = 0, then use the quadratic formula.
The second equation already isolates y as 16.83 - 2.49x. Substituting gives 19.6048x = 85.1016, so x = 4.3409 and y = 6.0213.
Answer: x >= 0, or [0, infinity). Cancel the 1 on both sides, combine -x/3 + 2x/7 over 21, and flip the sign when dividing by the negative coefficient.
The solution is x < 0. Combine -x - x into -2x, multiply both sides by the positive number 3 so the sign is unchanged, and collect the x terms.
The solution is x < 3/2. Combine 2x - x into x, multiply both sides by 3 which keeps the inequality direction, then collect terms to get 6x < 9.
The solution is all x not equal to 2. Factoring gives -(x-2)^2, which is negative everywhere except at its double root, where it is exactly zero and the strict inequality fails.
The answer is x < -|a| or x > |a|. You cannot take square roots directly because a may be negative; factoring into (x - a)(x + a) > 0 handles every sign of a at once.
Answer: 0 < x < 2945.715457. The left side is strictly increasing past x = 1 and negative below it, so a single numerical boundary settles the whole inequality.
The answer is x > 1. Taking logs turns it into x ln(x) > 0, and since x is positive the sign depends only on ln(x), which is positive exactly above 1.
The solution is x = 5. Multiplying every term by the least common denominator 6 clears all fractions in one move, turning the equation into 9x - 29 = x + 11.
Answer: x = 40000/59, about 677.97. Distribute 0.21, combine the x terms into 0.59x, subtract 2100 and divide; keeping the fraction avoids rounding drift.
The solution is 4 <= x <= 30. Factoring out the repeated bracket (800 - 20x) leaves (0.6 + 0.1x)(800 - 20x) >= 720, which rearranges to x^2 - 34x + 120 <= 0.
Answer: x = 142.393 or x = 17.619. The discriminant is 426213.1225 - 167051.52 = 259161.6025, whose square root is 509.0792; divide by 2a = 8.16 to finish.
Three real roots x ≈ −0.603957, 0.909281, 1.110778 plus the pair −0.070551 ± 0.636284i. No rational root exists, so the quintic must be solved numerically.
The five roots are x = 0, 10, −5, −15, −20. Factor out x, then split the quartic into (x² + 10x − 200)(x² + 20x + 75) — every root is an integer.
The only solution is x = (5 + sqrt33)/4, about 2.6861. Move one radical across so both sides are non-negative, then square twice and reject the extraneous root.
The solution is x = 162. Isolate the radical to get sqrt(x + 7) = 13 before squaring - squaring while the -11 is still attached is what produces the wrong answers.
The solutions are x = 0 and x = 1/16. Rewriting the nested radical as x^(3/4) turns the equation into x^(3/4) = 2x, and x = 0 must be handled separately.
On the domain x >= 1/3 both sides are non-negative, so squaring gives 3x - 1 > 4x^2, i.e. 4x^2 - 3x + 1 < 0. That quadratic has discriminant -7 and is always positive, so no x works.
The solution is x = 15. Rewriting the ratio as a fraction and cross-multiplying gives 15x + 6 = 14x + 21, so a single x survives on the left.
The solution is (2, -4). Multiply both equations by 10 to remove the decimals, scale by 2 and 3 to match the x coefficients, then add to get -23y = 92.
Answer: x = 6/95 and y = -6/47. Substituting a = 1/x and b = 1/y turns the reciprocal system into a linear one, solved by elimination and then inverted.
x = 3 and y = 3. Substituting y = 6 - x collapses the first equation to 120x + 600 = 960, and a weighted-average check confirms the even split of six units.
The answer is x = 4, y = -5. The first equation gives y = 3 - 2x with no fractions at all, which makes substitution cleaner than elimination here.
The solution is x = 14/5 and y = 9/5. Rearrange x − y = 1 into x = y + 1, substitute into 3x + 2y = 12 to get 5y + 3 = 12, then back-substitute.
Answer: x = 1 and y = 1. Scale the equations by 3 and 2 so the y terms become +6y and -6y, add to eliminate y, and 13x = 13 gives the answer immediately.
The solution is (23, -30). Multiply the first equation by 4 to get 2x + y = 16, substitute y = 16 - 2x into the second, and solve 2x - 48 = -2.
Multiply the first equation by 12 to get 8x - 9y = 6, expand the second to 2x + 7y = -17, then eliminate x to find y = -2 and x = -3/2.
The first inequality gives x <= 1 after dividing by -2 and flipping the sign; the second gives x > -7. The solution is the intersection (-7, 1].
The answer is x = 17/4 and y = 3/4. Dividing the second equation by 2 gives x + y = 5, after which subtracting eliminates x in a single step.
The solution is x = 3, y = -2. The y coefficients are already opposites, so adding the two equations eliminates y immediately and leaves 3x = 9 in one step.
The solution is x = 1, y = −2. Solving the first equation for y and substituting turns the pair into 5x + 2 = 7, a single linear equation.
In terms of s and p the system is s^2 - p = 7 and s + p = 5, giving s^2 + s - 12 = 0. Only s = 3, p = 2 gives real x and y, namely 1 and 2.
Since (x^2+y^2)^2 = x^4 + y^4 + 2(xy)^2 = 82 + 18 = 100, we get x^2 + y^2 = 10, then x + y = ±4 and x - y = ±2, giving the four real pairs (3,1), (1,3), (-3,-1), (-1,-3).
The solution is x = 5/3, y = 14/3. Both equations already give y, so set 4x - 2 = x + 3, solve for x, then substitute back into either line to get y.
The answer is x = 4. Add 3 to both sides to get (3/4)x = 3, then multiply by the reciprocal 4/3 rather than dividing by a fraction, or clear the 4 first.
The solutions are x = -2, 0 and 11. Move everything to one side, factor out the common x, then factor the remaining quadratic as (x - 11)(x + 2).
x = 5, a root of multiplicity three. The coefficients 1, -15, 75, -125 are the binomial pattern of (x - 5)^3, so there is one triple root and nothing else.
The only solution is x = 1, a triple root. The coefficients 1, −3, 3, −1 are the signed binomial row for the cube, so the cubic collapses to (x − 1)³ = 0.
The roots are x = 2 (a double root) and x = −1. Rational root testing finds x = 2, and synthetic division leaves x² − x − 2, which factors again to (x−2)(x+1).
The roots are x = 1, 2 and 3. Test the divisors of 6, divide out (x − 1) by synthetic division, factor x² − 5x + 6, then confirm with the Vieta sum and product.
The answer is about 737.56. Collecting the x terms gives 0.82x = 604.8, so the base is the fixed part divided by 1 minus the 18 percent share.
The answer is x = 600/7, about 85.71. Collecting the x terms gives 0.7x = 60, and the denominator 7 is exactly why the decimal never terminates.
The answer is x = 70/0.65 = 1400/13, about 107.69. The unknown sits on both sides, so collect the x terms to get 0.65x = 70 before dividing - not 70/0.35.
The answer is x = 400/3, about 133.33. Moving 0.4x over to the left gives 0.6x = 80, so the self-reference costs a factor of 1 - 0.4, not 0.4.
x = 262962.96. The unknown appears on both sides, so expand the brackets first; the equation collapses to 1.08x = 284000.
Answer: x = 37,500. The unknown sits inside the bracket too, so simplify to x = 0.12(350000 - x), expand to 1.12x = 42000, and divide to finish the problem.
The real roots are x = 1.0524735 and x = -1.6942810. The factored form x(x + 1)(x^2 + 1) guarantees exactly two real roots; the other two are complex.
The real roots are x = 1.0620313 and x = -1.7023038. Bracketing on [1, 1.1] and on [-1.75, -1.7] and then bisecting pins both down to seven decimals.
The two real roots are x = 7.9431476 and x = -8.4477787. Factoring the left side as x(x + 1)(x^2 + 1) shows immediately why exactly two real roots exist.
x = 7246.475 exactly. The left side collapses to 0.8x, which is the algebra of a 20 percent discount, and dividing by 0.8 is the same as multiplying by 1.25.
The answers are x = 0.4940 and x = -0.3340. Expanding gives x^2 - 0.16x - 0.165 = 0, whose discriminant 0.6856 is not a perfect square at all.
The answers are x = 4.143 and x = -3.983. The discriminant is 0.0256 + 66 = 66.0256, whose square root 8.1256 is nowhere near a round number.
The answer is x = 0.19/0.835 = 0.2275. This is the standard margin equation: a fixed cost of 0.19 leaves a 16.5% margin, so the base is cost over (1 - margin).
The answer is x = 50. Subtracting 0.98x leaves 0.02x = 1, and dividing by 0.02 is the same as multiplying by 50, since 0.02 is one fiftieth.
The answer is x = 100. Subtracting 0.99x leaves only 0.01x = 1, and dividing by the small coefficient 0.01 is the same as multiplying by 100.
Answer: x = 3. Multiply every term by the LCD 10 to get 5(x-1) = 20 - 2(x+2), expand both sides carefully, then collect terms to reach 7x = 21 in one move.
The answers are x = (3 + root 13)/2 and (3 - root 13)/2. Multiplying by x turns it into x^2 - 3x - 1 = 0, and neither root is zero, so neither is extraneous.
The exact answer is x = 512009/6 = 85334.83333…. Subtract 3, divide by 0.6 by multiplying by 5/3, then add 1; the repeating decimal signals a non-terminating exact value.
Answer: x = 168300/23, about 7317.39. The bracket expands to 0.08x - 500, so after distributing the minus sign the equation becomes 0.92x + 500 = 7232.
The bracket simplifies to x^2/(x+2), so with y = x^2/(x+2) the equation becomes y^2 + 4y - 5 = 0. Only y = 1 is usable, giving x = 2 or x = -1.
The solution is x <= 1. Combining x and -2x leaves a single -x, and dividing by -1 at the very last step reverses the inequality from >= into <=, giving x <= 1.
The answer is x > 8. Add 3 to both sides to isolate x; the inequality sign never changes because only addition, not a negative multiplier, is involved.
The solution is x > -3. Multiplying through by the LCD 4 gives x - 3 < 24 - 2(3 - 4x), and the minus in front of the fraction flips both inner signs.
Answer: x = 6500. Expand the bracket to 0.024x - 150, distribute the minus sign, and the equation collapses to 0.976x + 150 = 6494 with an exact whole-number root.
The solution is x <= 1. Distributing -3 across x - 2 gives -3x + 6, and the resulting -2x >= -2 flips direction when both sides are divided by -2.
The answer is x > 15. Add 4 to both sides to undo the subtraction; since no negative multiplication occurs, the greater-than sign is unaffected.
The solution is x > -5/4. Multiplying both sides by 6 gives 2(x - 4) < 3(2x - 1), and the answer stays an exact fraction rather than a decimal.
x = 7/2. Equal squares means the bases are equal or opposite; the 'equal' case gives x = 3.5 and the 'opposite' case collapses to the false statement -4 = -3.
The answer is x > 12. Adding 7 to both sides isolates x, and since addition never changes an inequality direction, the greater-than sign stays exactly as it is.
The root is x = -1. Both terms on the left contain x, so combine them into (4/5)x before dividing; multiplying through by 5 avoids fractions entirely.
The rearrangements are x = y + 8 and y = x - 8. One equation with two unknowns has infinitely many solutions, forming a line of slope 1 in the plane.
x = 1.9078533, exactly 17/3 times 0.33668. Cross-multiplying leaves 0.15x = 0.286178, and dividing by 0.15 produces a repeating decimal rather than a clean one.
x = 6.39692, which is exactly 19 times 0.33668. Cross-multiplying gives 0.05x = 0.319846, and the 5 percent remainder is what produces the factor 19.
The answer is x = 970/3, about 323.33. Cross-multiplying gives 3x = 970, and that 3 comes from 100 - 97 - which is why the decimal repeats forever.
The answer is x = 490. Cross-multiplying gives 50x = 49(10 + x), so x = 490 - a 98 percent target needs 49 times as much of x as of the fixed 10.
The answer is x < 2. Clear the fraction by doubling both sides, then collecting gives -x > -2, and multiplying by -1 reverses the direction to x < 2.
x = -20/3, about -6.6667. Multiplying every single term by 4 removes both denominators, giving 9x - 6 = 14 + 12x, and so -20 = 3x after collecting terms.
x = (-5 + sqrt 99545)/2 = 155.2538 or x = (-5 - sqrt 99545)/2 = -160.2538. Clearing the fifths and halves turns the product into x^2 + 5x - 24880 = 0.
The answer is x = 44/3, which is 14 and 2/3, or about 14.667. Multiply both sides by 8 to get 88/6, then reduce by the common factor of 2 to finish.
Clearing the denominators gives x^3 + 25x - 25x - 125 = 0, so x^3 = 125 and x = 5 is the only real root; the other two roots of the cubic are complex.
The answer is y = 100x/99, about 1.0101x. Multiplying by y gives x = 0.99y, and dividing by 0.99 turns the decimal into the exact fraction 100/99.
The answer is x = 2/3. Expanding both cubes gives 2x³ − 3x² + 15x − 7, and the subtracted term removes 2x³ − 3x², leaving the linear equation 15x − 7 = 3.
Answer: x = 13.15893318, exactly 92937321/7062679. Divide both sides by the eight-decimal multiplier, then subtract 1; rounding early shifts the result.
The solution is (3-sqrt(26))/2 <= x <= (3+sqrt(26))/2, roughly -1.0495 to 4.0495. Expand, move the 1/4 across, clear it, and use the quadratic formula on 4x^2-12x-17.
The solution is -1 < x < 4. A product of two factors is negative exactly when the factors have opposite signs, which happens strictly between the roots.
The solution is x > 2. Subtract x and then 6 to reach 4 < 2x, divide by the positive 2, and rewrite 2 < x with the variable on the left as x > 2.
The answer is x = 8. The four terms have coefficients 1, 10, 1 and 1, which add to 13, so the equation reduces to 13x = 104.
The answer is x < 4. Subtract 2 from both sides to isolate x; subtraction preserves the direction, so the less-than sign stays pointing the same way.
The solution is x < 17/2. Collecting the x terms on the right gives 17 > 2x, and dividing by 2 leaves the exact fraction 17/2 rather than a decimal.
The answer is x = 8. Both cubic terms cancel: the first product is the sum of cubes x³ + 64 and the second is x³ − 25x, leaving the linear equation 25x + 64 = 264.
The solutions are x = (-11 + sqrt(61))/2 and x = (-11 - sqrt(61))/2. Expanding gives x^2 + 11x + 15 = 0, whose discriminant 61 is not a perfect square.
Since x^2 + 1/x^2 = t^2 - 2, the equation becomes 2t^2 - 7t + 5 = 0 with t = 1 or 5/2. Only t = 5/2 has real preimages, giving x = 2 and x = 1/2.
The answer is x = 2. Since √4 = 2 is a constant, not a radical in x, the equation is ordinary linear: −2x + 2 = −2 gives x = 2 in two steps.
x = 1/28 and y = 1/21. Substituting x = 1/12 - y reduces the second equation to 2/3 + 7y = 1, giving y = 1/21; this is a classic two-rate work problem.
The answers are x = 1 + 3root 2 and x = 1 - 3root 2. The discriminant is 72, and simplifying root 72 to 6root 2 is what makes the fraction cancel neatly.
The answer is x = −8. Expanding the right side gives x² + 30x + 225; the x² terms cancel, leaving the linear equation 30x = −240.
The solutions are x = plus or minus sqrt(2 sqrt 3), about 1.8612. The right-hand side is irrational, so the answer is a nested radical equal to 2^(1/2) 3^(1/4).
x = 5 or x = -5. Recognising 25 as 5^2 turns the left side into (x - 5)(x + 5), which beats taking square roots because it shows both signs automatically.
The solutions are x = 1 + sqrt 5 and x = 1 - sqrt 5. The discriminant is 20, and simplifying sqrt 20 to 2 sqrt 5 lets the whole fraction cancel down neatly.
The solution is every real number except x = 1. The left side is the perfect square (x - 1)^2, which is strictly positive except at its repeated root x = 1.
The first gives (−∞, −2] ∪ [5, ∞) and the second gives [−1, 1]; their intersection is empty, so no single x satisfies both quadratic inequalities at once.
The solution is -2 <= x <= 5. The factorisation (x - 5)(x + 2) shows the upward parabola dips below the axis only between its roots, endpoints included.
The answers are x = 1 and x = 3. The numbers -1 and -3 multiply to give 3 and add to give -4, so the quadratic factors cleanly as (x - 1)(x - 3).
The answer is y = (x^2 + 1)/4. Isolate the -4y term, then divide by -4 so both signs flip; the result is an upward parabola with vertex at (0, 1/4).
The solutions are x = 2 and x = 3. A positive constant with a negative middle term forces both factors negative, and -2 times -3 is 6 while -2 plus -3 is -5.
The answers are x = 3 + root 14 and 3 - root 14. The discriminant is 56, and simplifying root 56 to 2root 14 lets the factor of 2 cancel with the denominator.
The solution is x < (7 - sqrt 73)/2 ≈ -0.7720 or x > (7 + sqrt 73)/2 ≈ 7.7720. The discriminant 73 is prime, so both boundaries remain irrational numbers.
The answer splits on |a| > 2: the solution is the open interval between (a - root(a^2-4))/2 and (a + root(a^2-4))/2, and is empty when -2 <= a <= 2.
x = 7 or x = -6. Look for two numbers with product -42 and sum -1; they are -7 and 6, giving the factorisation (x - 7)(x + 6) = 0.
x = (-17 + sqrt(769))/2 ≈ 5.3654 or x = (-17 - sqrt(769))/2 ≈ -22.3654. The discriminant 769 is not a perfect square, so this quadratic has no integer factoring.
The answers are x = -1 + root 2 and -1 - root 2. Rewriting as (x + 1)^2 = 2 is faster than the formula here, because the middle coefficient is even.
The solutions are x = -1 and x = -2. Adding 2 to both sides gives x^2 + 3x + 2 = 0, which factors as (x + 1)(x + 2) because 1 times 2 is 2 and 1 plus 2 is 3.
x = -6 or x = 3. Two numbers with product -18 and sum +3 are 6 and -3, so the quadratic factors as (x + 6)(x - 3) = 0.
x = (-3 +/- i sqrt 11)/2. Subtracting 1 gives x^2 + 3x + 5 = 0, whose discriminant is -11, so there are no real roots at all - only a conjugate pair.
The solutions are x = -7 and x = 3. A negative constant term means the factor pair has opposite signs, and 7 times -3 is -21 while 7 plus -3 is 4.
x = -2 or x = -3. Two numbers that multiply to 6 and add to 5 are 2 and 3, so the quadratic factors, and the zero product property then gives both roots.
x = 14 or x = -15. Moving 210 across gives x^2 + x - 210 = 0, and the pair 15 and -14 multiplies to -210 while adding to 1, which is the x coefficient.
The solution is x <= (-1 - sqrt 5)/2 ≈ -1.6180 or x >= (-1 + sqrt 5)/2 ≈ 0.6180, and those two boundary values are exactly the golden ratio conjugates.
The answer is x = 5180/11 = 470.909091, not the round 470. Dividing 518 by 1.10 never terminates, and 1.1 x 470 = 517 falls a full unit short of 518.
The answer is x = 7/12, about 0.5833. Divide both sides by 60 to get 35/60, then cancel the common factor of 5 from numerator and denominator.
There is no real solution: the discriminant of x^2 - 37x + 345 is -11. The complex answers are (37 plus or minus i root 11)/2, so no rectangle of that area exists.
x = 1828/365 = 5.00822, not exactly 5. Collapsing 75 * 365 / 1000 to 27.375 gives 27.375x = 137.1, while 27.375 * 5 is only 136.875 - short by 0.225.
The answers are x = 15 and x = 35. Dividing through by -5 turns the equation into x^2 - 50x + 525 = 0, which factors neatly as (x - 15)(x - 35).
The four roots are x = -2, -sqrt 2, sqrt 2 and 2. The substitution u = x^2 turns the quartic into u^2 - 6u + 8 = 0, which factors as (u - 2)(u - 4).
The solution is x = 10^25. Setting y = log x turns the equation into y^(3/2) = 125, so y = 125^(2/3) = 25 and x = 10^25.
The unique real root is x ≈ 1.038016. The left side is strictly increasing, so exactly one root exists; f(1) = 4 and f(1.1) = 5.990 bracket it.
Dividing by -3x flips the sign of every term: 3x^2 becomes -x, -15x becomes +5, and the constant 5 becomes -5/(3x).
Divide each numerator term by 4x^2 separately: subtract exponents and divide the coefficients. Only the middle term keeps a fraction, as 5/4.
The decomposition is 1/(x - 2) + 2/(x - 2)^2. Writing x as (x - 2) + 2 lets each piece cancel against the denominator with no simultaneous equations at all.
The product is -19. Writing root(-19) = i root 19 first is essential, because the rule root a times root b = root(ab) fails for negative radicands.
The expansion is 3a^4 + 24a^3 + 48a^2. Factoring root 3 a out first turns the problem into 3a^2(a + 4)^2, and squaring removes the radical entirely.
A + C has rows (19,3,3), (3,5,3), (3,37,3) and its square has rows (379,183,75), (81,145,33), (177,305,129). Add entrywise, then multiply the sum by itself.
The quartic has just two real roots, x ≈ −78.5149849 and x ≈ 80.4985052. Between them the quartic never rises above about −3992, so no root exists anywhere in [0, 80].
The difference is -3x^3 + 6x^2 + 7x - 18 and its degree is 3. Distribute the minus sign over the second polynomial first, then combine like terms.
The difference is 41/((x - 2)(x - 6)). Factoring the quadratic denominator reveals the common denominator, and the x-terms cancel to leave a constant numerator.
The difference is −6x⁴y − 2x³y² + 9x²y³ − 3xy⁴ + y⁵. Learn how to negate every term of the second bracket and which two-variable terms actually combine.
Answer: -4 + 15i. Distribute the minus sign over both parts of the second complex number, then add real parts and imaginary parts separately.
The difference equals −2y times the cube root of 2x. Pull the perfect cubes 8 and y³ out of each radical so both terms become like radicals in cube root of 2x.
The difference is n^3 + 2. Distribute the minus sign across the whole second bracket, watch the two n^7 terms cancel, and combine the remaining like terms.
The answer is −2x² − 4x + 62. The whole difficulty is distributing the minus sign across all three terms of g, which flips 4x to −4x and −60 to +60.
Answer: (-x^2 - 4x + 14)/((x+5)(x+4)). Cross-multiply onto the common denominator, expand both products, and distribute the minus sign across all three terms.
The difference is x^4y^2 - 15x^3y - 14y + 8x with degree 6. Distribute the minus sign across all four terms, combine like terms, then add exponents per term.
The difference is -x^3 - 10xy - 2y^2, a polynomial of degree 3. Distribute the minus sign, then pair the x^3, xy and y^2 terms before reading off the highest total degree.
The sum is 10s^2t - 5st^2, a binomial with a degree of 3. Only like terms combine, and each term's degree is the sum of its exponents, 2 + 1 = 3, not 2.
The system u3 + u5 - u6 = 6 and u8 + u4 = 52 gives d = 5 and u1 = 1. The even-indexed terms form an AP with difference 10, and their 1010-term sum is 5101510.
Answer: the sum is 6 (m = 0, 2, 4). The system is empty when m >= -1, and x + m - 2 = 2 - x has a non-negative integer root when m is even and m <= 4.
The answer is 60, from the two valid first terms 24 and 36. The nine angles total 540, so the middle term is 60, and 78 forces (k - 5)d = 18 with d dividing 18.
The sum of the roots is -3. The domain x >= -5 throws out x = -6, while the radical factor itself contributes the extra root x = -5, so the roots are 2 and -5.
The sum is -3. The domain x >= -5 rejects the quadratic root x = -6, leaving x = 2 from the quadratic factor and x = -5 from the square-root factor.
The sum is 3. The domain is [-1, 4], so the candidate x = 5 is rejected and only x = -1, x = 0 and x = 4 survive the zero-product test on the two factors.
The two real roots are x ≈ 0.027167 and x ≈ −2.122636. Substituting y = 1 + x makes the left side a geometric series, y²⁷(y¹⁰ − 1)/(y − 1), which is strictly increasing only for y > 0 — the negative branch hides a second root.
The solution is the interval -3/2 <= x <= 3/2. Splitting at x = -1 and x = 1 turns the sum into -2x, then the constant 2, then 2x; only the outer cases bind.
Rationalising gives x = 2 + √3 and y = 2 − √3, so x + y = 4 and xy = 1; the symmetric identities then give x² + y² = 14, x³ + y³ = 52 and 4x² + 7xy + 4y² = 63.
Rationalising gives x = sqrt5 + 2 and y = sqrt5 - 2, so the sum is 2 sqrt 5 and the product is 1. Then x^2 + y^2 = (x+y)^2 - 2xy = 20 - 2 = 18.
Dividing gives quotient −x²/2 − x/4 + 27/8 and remainder 21/8. Factor the divisor as −2(x − 1/2) first, divide by x − 1/2, then divide the quotient by −2.
The quotient is 2t² + 2t + 1 and the remainder is −14. The missing t² term must be entered as a 0 placeholder, giving the coefficient row 2, 0, −1, −15.
The quotient is 2t + 1 and the remainder is −14, so 2t² − t − 15 = (t − 1)(2t + 1) − 14. A leading coefficient of 2 is fine; only the divisor must be monic.
The quotient is 2x^2 + 4x + 6 with remainder 5. The divisor x - 1 gives c = 1, and the remainder theorem confirms the 5 independently.
Synthetic division gives quotient x² + 2x + 2 and remainder 7. See how the 0 placeholder for the missing x² term and the bring-down rows work, row by row.
The quotient is x² − 4 with remainder 0, so x − 2 divides the cubic exactly. Set c = 2, bring down 1, then multiply-and-add through the coefficients.
The quotient is x² + 2x and the remainder is 5, so the cubic equals (x − 2)(x² + 2x) + 5. Pad the missing x² term with 0 and use c = 2 in the table.
The quotient is x² − 5x + 6 with remainder 0, so the cubic factors as (x + 1)(x − 2)(x − 3). Note that x + 1 means c = −1, not c = 1, in the table.
The quotient is x² + x + 2 and the remainder is 0. The cubic has no x term, so a placeholder 0 must go in that column before the table is run.
The quotient is x² + 4x − 2 and the remainder is −10, so the cubic equals (x − 1)(x² + 4x − 2) − 10. A nonzero remainder means x − 1 is not a factor.
The solution is x = 300, y = 140. Dividing the equations by 8 and 6 gives x + y = 440 and x + 2y = 580; subtracting isolates y immediately.
Answer: x = 35, y = 5. Expand both brackets to reach 3x - 2y = 95 and 2x + y = 75, solve the second for y, and substitute to get 7x = 245 in one line.
There is no real solution. The first inequality gives -3 < x < 2 and the second gives x < -6 or x > 2, and those two solution sets never overlap.
z = 14/13 and the pairs are ((28 ∓ √595)/13, (14 ± √595)/13). The trick is spotting that the middle equation rearranges into 2 × the third equation, which pins z immediately.
Adding the equations gives 2y = 12, so y = 6; substituting back forces x^2 = -4, which no real number satisfies. The system has no real solution.
The third alloy weighs 150 kg. Balance the copper: 0.40(100) + 0.55m = 0.45(100 + m), which gives m = 50 kg of the second alloy on top of the first 100 kg.
The integers are 16, 17 and 18. Calling them n-1, n, n+1 makes the sum exactly 3n, so n = 51/3 = 17 and the middle number falls out immediately.
One of each costs 6.00. Two white plus one yellow is 5, two white plus three red is 10.50, and three yellow plus two red is 11 - solve it by substitution.
The system is inconsistent. Eliminating b and c makes every a cancel and leaves 2 = 100, a false statement, so no triple (a, b, c) can satisfy all three.
The solution is x ≈ 0.21936, y ≈ −0.14766, z ≈ 0.07169. Substituting z = x + y first turns three equations into two, and reading 25.33333 as 76/3 gives exact fractions.
The row sizes form an AP with u1 = 15, d = 4, n = 30, so the last row holds 131 seats and the total is (30/2)(15 + 131) = 2190 seats.
The combined volume is 16.6xy cubic metres, which is 249 cubic metres at x = 5 and y = 3. Tripling both base dimensions multiplies that base area by 9, not by 3.
The answer is 77 km/h. Writing the distance as 10v makes the leftover leg exactly v kilometres, so the delay equation reduces to v/(v − 7) = 11/10.
The result is f(x) = 2|x + 5| − 8. A shift right by 1 replaces x with x − 1, moving the vertex from (−6, −1) to (−5, −1); the drop of 7 then lowers it to (−5, −8).
The result is f(x) = |x − 3| + 9 and the vertex moves from (1, 5) to (3, 9). Both constants change: 1 becomes 3 inside the bars and 5 becomes 9 outside them.
The result is f(x) = |x − 6| + 3, with the vertex moving from (4, −3) to (6, 3). Horizontal shifts go inside the bars and subtract; vertical shifts go outside and add.
The result is f(x) = |x − 6| + 6, moving the vertex from (4, 0) to (6, 6). With no constant term to start, the up-shift supplies the outside constant directly.
Replacing x with x + 4 gives |(x + 4) - 5| = |x - 1|, moving the vertex from x = 5 to x = 1. A left shift uses a plus sign inside the function.
A vertical shift subtracts outside the absolute value, giving |x + 2| - 3. The vertex drops from (-2, 0) to (-2, -3), and the graph now crosses the x-axis twice.
The result is f(x) = |x + 8| + 8. A horizontal shift goes inside the absolute value with the opposite sign; a vertical shift goes outside with the same sign.
The result is f(x) = |x − 2| + 4 with vertex (2, 4). Starting from the parent |x| makes the rule visible: h goes inside with a minus sign, k goes outside with a plus.
There is no choice of q, r, s. With p = 1 the first conjunct reduces to r ∧ ¬s and the second to s ∨ ¬r, and those two demands contradict each other, so the formula is false for every option.
The speeds are 12 km/h and 15 km/h. Setting 120/x − 120/(x+3) = 2 clears to x² + 3x − 180 = 0, whose positive root is x = 12.
Answer: x = 3/5 and y = 9/5. The second equation gives x = y/3 with no fractions to clear, and substituting turns the first into (10/3)y = 6 in one step.
There is no solution. Both expressions have slope 0.055, so their difference is the constant 2440 for every x — the graphs are parallel lines that never meet.
The answer is 4 days. Two days alone leaves 3/5 of the house, and the combined rate of 3/10 per day clears that remainder in exactly 2 more days.
The line is y = −3x + 689. The equation is two-point form for the points (362, −397) and (366, −409); cross-multiplying and dividing by 4 gives slope −3.
One of each costs 6 pounds: white 1.50, yellow 2.00, red 2.50. Substituting y = 5 - 2w into the third equation reduces the system to a single equation in w.
Each tin costs 18 dollars. Three friends paid equally, so the fair share is the average; the first friend took 8 tins above it, and 144 divided by 8 gives the price.
The value is 2. Vieta gives x₁ + x₂ = −3 and x₁x₂ = −5, so the expression equals −3 − (−5) = 2 without ever finding the irrational roots.
Every real m works. Rearranging gives m ≤ x² − x + 1, and on x > 0 that expression is unbounded above, so a large enough positive x always satisfies the inequality.
The balance after n weeks is 42 + 8(n - 1) = 8n + 34. Solving 8n + 34 >= 400 gives n >= 45.75, so week 46 with 402 dollars is the first that suffices.
The condition says z^2025 equals its own conjugate, so z^2025 is real. In polar form it forces arg(z) = k*pi/2025 for an integer k, or z = 0, with the modulus free.
If n is odd, a + b is always a factor with alternating signs in the cofactor. If n is even there is no general factorisation, though n = 6 works via cubes.
The critical points 0, 2/3 and 1 split the line into four intervals; testing one point in each shows the fraction is positive on (-inf, 0) and (2/3, 1).
Option C is the only correct one. A gives a non-constant 'minimum', B's minimum is not attained on x > 1, and D's value 1 is a maximum, not a minimum, because 4x - 5 is negative.
A step function. Because each additional ounce or any part of one costs a full 0.20, the total jumps at whole-ounce boundaries and stays flat in between.
A and C are correct. The product ab ranges over (-4, 2), not (-2, 2), and ab - 2 < a - 2b is false because it rearranges to (a + 2)(b - 1) < 0 with both factors positive.
Parallel lines have equal slopes. Reading m from y = mx + b, only y = 4x + 7 has slope 4; the -1/4 option is perpendicular, not parallel.
Only statement 2 is correct. Rewrite every difference as A - B = A ∩ B', apply De Morgan's laws, and compare: the other three fail because union and complement do not commute that way.
B and C are correct. xy has maximum 1/4 (not minimum), and sqrt(x) + sqrt(y) has maximum sqrt(2), not 2, because its square is 1 + 2 sqrt(xy) <= 2.
From u_5 - u_2 = 3d we get d = 6 and u_1 = -5. Solving -5 + 6(n - 1) = 103 gives n = 19, and u_19 = -5 + 108 = 103 confirms it.
All three coefficients are multiples of 5, so the left side is always divisible by 5 — but 21,806 leaves remainder 1. Over the reals the solution set is a plane.
The discriminant is -16, which is negative, so there are no real roots and no real linear factors. Over the complex numbers it splits as 5(x - r)(x - r-bar).
Yes: subtracting uₙ₋₁ = u₁ + (n−2)d from uₙ = u₁ + (n−1)d leaves exactly d. The identity characterises arithmetic sequences and fails for any other kind.
Neither 1 nor -1 is a root, so it is irreducible over the rationals. Its derivative 3x^2 - 6x + 5 is always positive, so the cubic is strictly increasing.
All four candidates from the Rational Root Theorem fail, so it is irreducible over the rationals - yet it still has three real roots, near -0.391, 1.227 and 4.164.
x = 3 is excluded because it makes the denominator (x−3)(x+3) zero, giving the undefined form 0/0. Cancelling x − 3 leaves a hole at x = 3, not a valid value.
It does not factor. Read as a quadratic in x, its discriminant is 16y(y - 1), which is not a perfect square polynomial, unlike the familiar x^2 - 4xy + 4y^2.
256/625 = (4/5)^4, because 256 = 4^4 and 625 = 5^4. Since both parts share the exponent 4, the fraction folds into one power of 4/5, or 2^8/5^4.
Answer: 81/16 = (3/2)^4. Factor 81 = 3^4 and 16 = 2^4, then use a^n/b^n = (a/b)^n. As a decimal 81/16 = 5.0625, which is no integer power of a whole number.
The answer is x + 66 when x >= -66, and -x - 66 when x < -66. The split point is where the inside changes sign, found by solving x + 66 = 0 for x.
The answer is x^2 for x >= 0 and -x^2 for x < 0. Splitting at zero removes the absolute value, and the result is an odd, everywhere-increasing function.
The equation is 2((2w + 2) + w) = 64, giving width 10 m and length 22 m. The key is turning 'two more than twice the width' into 2w + 2, not 2(w + 2).
The equation is 2((3w + 4) + w) = 96, which solves to width 11 m and length 37 m. Same translation pattern as the 2w + 2 version but with a different multiplier.
The answer is -4x + 4y + 5. A minus sign in front of a bracket multiplies every term inside by -1, so all three signs flip - including the last one.
The answer is x = 6. Setting the equations equal gives 2x^2 - 24x + 64 - a = 0, and one intersection means a zero discriminant, so x = 24/4 = 6.