इस पर अटक गए? इसे AI से हल करें
Algebra · real student question

Solve the cubic equation 2s^3 + 33s^2 + 180s + 300 = 0.

Question

Solve

2s3+33s2+180s+300=02s^3 + 33s^2 + 180s + 300 = 0

Step-by-step solution

  1. Rule out rational roots. Candidates are ±de\pm\tfrac{d}{e} with d∣300d \mid 300 and e∣2e \mid 2. Testing the plausible negatives: f(−3)=3f(-3) = 3, f(−4)=−20f(-4) = -20, f(−5)=−25f(-5) = -25, f(−6)=−24f(-6) = -24, f(−10)=−200f(-10) = -200, f(−15)=−1725f(-15) = -1725. None vanish, so factoring and synthetic division are both dead ends — the sign change between −3-3 and −4-4 does tell us one real root lies in that gap.

  2. Make the cubic monic and depress it. Dividing by 22 gives s3+332s2+90s+150=0s^3 + \tfrac{33}{2}s^2 + 90s + 150 = 0. The substitution s=y−a3s = y - \tfrac{a}{3} with a=332a = \tfrac{33}{2}, i.e. s=y−112s = y - \tfrac{11}{2}, removes the square term. Using p=b−a23p = b - \tfrac{a^2}{3} and q=2a327−ab3+cq = \tfrac{2a^3}{27} - \tfrac{ab}{3} + c:

    p=90−108912=−34,q=13314−345=−494p = 90 - \frac{1089}{12} = -\frac{3}{4}, \qquad q = \frac{1331}{4} - 345 = -\frac{49}{4}

    so the depressed cubic is y3−34y−494=0y^3 - \tfrac{3}{4}y - \tfrac{49}{4} = 0.

  3. Compute the discriminant to see how many real roots there are.

    Δ=(q2)2+(p3)3=240164−164=240064=752=37.5\Delta = \left(\frac{q}{2}\right)^2 + \left(\frac{p}{3}\right)^3 = \frac{2401}{64} - \frac{1}{64} = \frac{2400}{64} = \frac{75}{2} = 37.5

    Since Δ>0\Delta > 0 there is exactly one real root and a conjugate pair of complex roots. This is the case where Cardano's radical formula works with real cube roots — no trigonometric detour needed.

  4. Apply Cardano's formula. With −q2=498=6.125-\tfrac{q}{2} = \tfrac{49}{8} = 6.125 and Δ=37.5=6.123724\sqrt{\Delta} = \sqrt{37.5} = 6.123724:

    y=6.125+6.1237243+6.125−6.1237243=12.2487243+0.0012763y = \sqrt[3]{6.125 + 6.123724} + \sqrt[3]{6.125 - 6.123724} = \sqrt[3]{12.248724} + \sqrt[3]{0.001276}

    The two radicands are wildly different in size — the second is barely above zero because Δ\Delta falls just short of (q2)2\left(\tfrac{q}{2}\right)^2 — giving

    y=2.305138+0.108453=2.413591y = 2.305138 + 0.108453 = 2.413591

  5. Shift back to s and find the complex pair. With s=y−112s = y - \tfrac{11}{2} the real root is

    s1=2.413591−5.5=−3.086409s_1 = 2.413591 - 5.5 = -3.086409

    Dividing 2s3+33s2+180s+3002s^3 + 33s^2 + 180s + 300 by (s−s1)(s - s_1) leaves 2s2+26.827182s+97.2003442s^2 + 26.827182s + 97.200344, whose roots are s=−6.706796±1.902385 is = -6.706796 \pm 1.902385\,i.

  6. Check with Vieta. The three roots must sum to −332=−16.5-\tfrac{33}{2} = -16.5: −3.086409+2(−6.706796)=−16.500001-3.086409 + 2(-6.706796) = -16.500001. Their product must be −3002=−150-\tfrac{300}{2} = -150: (−3.086409)(6.7067962+1.9023852)=(−3.086409)(48.60018)=−150.00(-3.086409)\left(6.706796^2 + 1.902385^2\right) = (-3.086409)(48.60018) = -150.00. Both match. (A commonly seen but wrong answer gives the imaginary part as ±0.484891\pm 0.484891; that pair fails the product test badly.)

Answer

s≈−3.086409,s≈−6.706796±1.902385 is \approx -3.086409, \qquad s \approx -6.706796 \pm 1.902385\,i

Need to solve a different problem like this? Open the solver →