Multiple choice

If $p$ and $q$ are non-zero real numbers and $\alpha^{3}+\beta^{3}=-p, \alpha \beta=q$, then a quadratic equation whose roots are $\dfrac{\alpha^{2}}{\beta}, \dfrac{\beta^{2}}{\alpha}$ is :

  1. $px^{2}-qx+p^{2}=0$
  2. $qx^{2}+px+q^{2}=0$
  3. $px^{2}+qx+p^{2}=0$
  4. $qx^{2}-px+q^{2}=0$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Given alpha^3 + beta^3 = -p and alpha*beta = q. We know alpha^3 + beta^3 = (alpha+beta)(alpha^2 - alpha*beta + beta^2) = (alpha+beta)((alpha+beta)^2 - 3*alpha*beta). Let S = alpha+beta. S(S^2 - 3q) = -p. The roots of the new equation are a = alpha^2/beta and b = beta^2/alpha. Sum = (alpha^3 + beta^3)/(alpha*beta) = -p/q. Product = alpha*beta = q. The equation is x^2 - (sum)x + product = 0, so x^2 + (p/q)x + q = 0, which is qx^2 + px + q^2 = 0.

AI explanation

To find the sum of the new roots, calculate alpha^2/beta + beta^2/alpha, which equals (alpha^3 + beta^3)/(alpha*beta). Substituting the given values alpha^3 + beta^3 = -p and alpha*beta = q gives the sum as -p/q. The product of the new roots is (alpha^2/beta)*(beta^2/alpha), which equals alpha*beta = q. Using these, the new quadratic equation is x^2 - (sum)x + product = 0, resulting in x^2 + (p/q)x + q = 0, and multiplying by q gives the final equation qx^2 + px + q^2 = 0.

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