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Difficulty: Very hardQuadratic Equations and Factoring

The quadratic equation x2−px+q=0x^2 - px + q = 0, where pp and qq are real constants with p>0p > 0, has two real roots r1r_1 and r2r_2. If r12+r22=29r_1^2 + r_2^2 = 29 and r13+r23=117r_1^3 + r_2^3 = 117, what is the value of qq?

Answer: -10

Answer

The value of qq is −10-10.
By Vieta's formulas, r1+r2=pr_1 + r_2 = p and r1r2=qr_1 r_2 = q. Using the identity r12+r22=(r1+r2)2−2r1r2=29r_1^2 + r_2^2 = (r_1 + r_2)^2 - 2r_1 r_2 = 29, we get q=p2−292q = \frac{p^2 - 29}{2}. Using the sum of cubes identity r13+r23=(r1+r2)(r12−r1r2+r22)=117r_1^3 + r_2^3 = (r_1 + r_2)(r_1^2 - r_1 r_2 + r_2^2) = 117, we obtain p(29−q)=117p(29 - q) = 117. Substituting qq into this equation yields p3−87p+234=0p^3 - 87p + 234 = 0. Factoring this cubic equation yields p=3p = 3 as a valid root, leading to q=−10q = -10. Evaluating the discriminant p2−4q=9−4(−10)=49>0p^2 - 4q = 9 - 4(-10) = 49 > 0 confirms that real roots exist. The second positive root for pp gives a negative discriminant, making −10-10 the unique correct value of qq.

Step-by-Step Solution

1
Express symmetric sums of roots in terms of equation coefficients pp and qq using Vieta's formulas.
r1+r2=pr_1 + r_2 = p and r1r2=qr_1 r_2 = q. Therefore, r12+r22=(r1+r2)2−2r1r2=p2−2q=29r_1^2 + r_2^2 = (r_1 + r_2)^2 - 2r_1 r_2 = p^2 - 2q = 29, which gives q=p2−292q = \frac{p^2 - 29}{2}.
Relating root sums and products directly to coefficients simplifies the algebraic system.
2
Express the sum of cubes in terms of pp and qq and substitute qq.
r13+r23=(r1+r2)(r12−r1r2+r22)=p(29−q)=117r_1^3 + r_2^3 = (r_1 + r_2)(r_1^2 - r_1 r_2 + r_2^2) = p(29 - q) = 117. Substituting q=p2−292q = \frac{p^2 - 29}{2} yields p(29−p2−292)=117  ⟹  p(87−p2)=234  ⟹  p3−87p+234=0p\left(29 - \frac{p^2 - 29}{2}\right) = 117 \implies p(87 - p^2) = 234 \implies p^3 - 87p + 234 = 0.
Combining symmetric expressions forms a single polynomial equation in variable pp.
3
Factor the cubic equation p3−87p+234=0p^3 - 87p + 234 = 0 to find candidate values for p>0p > 0.
Testing factor p=3p = 3 gives 33−87(3)+234=27−261+234=03^3 - 87(3) + 234 = 27 - 261 + 234 = 0. Factoring out (p−3)(p - 3) gives (p−3)(p2+3p−78)=0(p - 3)(p^2 + 3p - 78) = 0.
Finding the roots of the polynomial identifies candidate values for coefficient pp.
4
Verify the real root constraint p2−4q>0p^2 - 4q > 0 for candidates of pp.
For p=3p = 3, q=32−292=−10q = \frac{3^2 - 29}{2} = -10, giving discriminant p2−4q=9−4(−10)=49>0p^2 - 4q = 9 - 4(-10) = 49 > 0 (real roots 55 and −2-2). For the positive root p=321−32≈7.458p = \frac{\sqrt{321}-3}{2} \approx 7.458 of p2+3p−78=0p^2 + 3p - 78 = 0, the discriminant is p2−4q=3p−29≈−6.63<0p^2 - 4q = 3p - 29 \approx -6.63 < 0, which yields non-real complex roots.
The question specifies real roots, so candidate solutions resulting in negative discriminants must be discarded.

Key Concept

Quadratic Equations and Factoring (Vieta's Formulas, Symmetric Polynomials, and Real Root Conditions)
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Topics

Quadratic Equations and Factoring

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