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Algebra · real student question

Solve (515 − 0.27x)(368 − 0.27x)(98 − 0.18x) − (515 − 0.27x)·98² − (98 − 0.18x)·270² = 0 for x.

Question

Solve for xx:

(515−0.27x)(368−0.27x)(98−0.18x)−(515−0.27x)⋅982−(98−0.18x)⋅2702=0.(515-0.27x)(368-0.27x)(98-0.18x)-(515-0.27x)\cdot 98^{2}-(98-0.18x)\cdot 270^{2}=0.

Step-by-step solution

  1. Abbreviate the three linear factors. Put

    A=515−0.27x,B=368−0.27x,C=98−0.18x,A=515-0.27x,\qquad B=368-0.27x,\qquad C=98-0.18x,

    so that, using 982=960498^{2}=9604 and 2702=72900270^{2}=72900, the equation is

    ABC−9604A−72900C=0.ABC-9604A-72900C=0.

    Naming the factors keeps the bookkeeping manageable and makes the common factor AA visible.

  2. Expand BCBC and group.

    BC=(368−0.27x)(98−0.18x)=36064−92.7x+0.0486x2,BC=(368-0.27x)(98-0.18x)=36064-92.7x+0.0486x^{2},

    so BC−9604=26460−92.7x+0.0486x2BC-9604=26460-92.7x+0.0486x^{2} and the equation becomes

    (515−0.27x)(0.0486x2−92.7x+26460)−72900(98−0.18x)=0.(515-0.27x)\bigl(0.0486x^{2}-92.7x+26460\bigr)-72900(98-0.18x)=0.

  3. Expand fully with exact fractions. Working with 0.27=271000.27=\tfrac{27}{100}, 0.18=181000.18=\tfrac{18}{100} rather than decimals avoids drift; the result is

    −0.013122x3+50.058x2−41762.7x+6482700=0,-0.013122x^{3}+50.058x^{2}-41762.7x+6482700=0,

    Multiplying the original by −5000006561-\tfrac{500000}{6561} gives the monic cubic x3−10300027x2+2320150000729x−120050000000243=0x^{3}-\tfrac{103000}{27}x^{2}+\tfrac{2320150000}{729}x-\tfrac{120050000000}{243}=0.

  4. Find the one exact root. Testing values that make a bracket round, x=2450027x=\tfrac{24500}{27} makes 0.27x=2450.27x=245 exactly, and substituting it into the cubic with exact fractions gives 00. So (x−2450027)\left(x-\tfrac{24500}{27}\right) is a factor and x=2450027=907.407‾x=\tfrac{24500}{27}=907.\overline{407} is a root.

  5. Divide out and solve the quadratic. Synthetic division leaves x2−7850027x+49000009=0x^{2}-\tfrac{78500}{27}x+\tfrac{4900000}{9}=0, i.e.

    27x2−78500x+14700000=0,27x^{2}-78500x+14700000=0,

    with discriminant 785002−4(27)(14700000)=457465000078500^{2}-4(27)(14700000)=4574650000. Since 4574650000=104⋅4574654574650000=10^{4}\cdot 457465 and 457465=5⋅91493457465=5\cdot 91493 is square-free, the roots stay irrational:

    x=39250±5045746527.x=\frac{39250\pm 50\sqrt{457465}}{27}.

  6. State the three roots and verify. Numerically x≈201.1822413x\approx 201.1822413, x=907.4074074x=907.4074074 and x≈2706.2251661x\approx 2706.2251661. Substituting each back into the original product form gives 00, −9×10−10-9\times10^{-10} and 00 respectively — the tiny residual is floating-point noise on a value of order 10610^{6}. A careless expansion that produces −41777.7x+6479700-41777.7x+6479700 instead is off: that cubic has no root at 24500/2724500/27.

Answer

x=2450027≈907.4074,x=39250−5045746527≈201.1822,x=39250+5045746527≈2706.2252x=\frac{24500}{27}\approx 907.4074,\quad x=\frac{39250-50\sqrt{457465}}{27}\approx 201.1822,\quad x=\frac{39250+50\sqrt{457465}}{27}\approx 2706.2252

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