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Zorluk: Çok zorQuadratic 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?

Cevap: -10

Cevap

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.

Adım Adım Çözüm

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.

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