Multiple choice

If the roots $\alpha, \beta$ of the equation $px^{2} + qx + r = 0$ are real and of opposite sign (where $p, q, r$ are real coefficient), then the roots of the equation $\alpha (x - \beta)^{2} + \beta (x - \alpha)^{2} = 0$ are

  1. Positive

  2. Negative

  3. Real and of opposite sign

  4. Imaginary
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Given roots alpha and beta of px^2 + qx + r = 0 are real and opposite sign. The equation alpha(x-beta)^2 + beta(x-alpha)^2 = 0 expands to (alpha+beta)x^2 - 4*alpha*beta*x + alpha*beta(alpha+beta) = 0. The discriminant D = 16*alpha^2*beta^2 - 4(alpha+beta)^2*alpha*beta. Since alpha and beta have opposite signs, alpha*beta < 0. Thus, D > 0, meaning roots are real. The product of roots is alpha*beta(alpha+beta)/(alpha+beta) = alpha*beta < 0, so roots are of opposite sign.

AI explanation

Expanding the equation alpha(x - beta)^2 + beta(x - alpha)^2 = 0 gives (alpha + beta)x^2 - 4*alpha*beta*x + alpha*beta(alpha + beta) = 0. The discriminant of this new quadratic is 16*alpha^2*beta^2 - 4*alpha*beta(alpha + beta)^2, which factors to -4*alpha*beta(alpha - beta)^2. Because the original roots alpha and beta are real and of opposite sign, their product alpha*beta is negative, making the discriminant positive and ensuring the new roots are also real and of opposite sign.

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