Multiple choice

Let $p$ and $q$ be real numbers such that $p\neq 0,\ p^{3}\neq q$ and $p^{3}\neq-q$. If $\alpha$ and $\beta$ are non zero complex numbers satisfying $\alpha+\beta=-p$ and $\alpha^{3}+\beta^{3}=q$, then a quadratic equation having $\dfrac{\alpha}{\beta}$ and $\dfrac{\beta}{\alpha}$ as its roots is

  1. $(p^{3}+q)x^{2}-(p^{3}+2q)x+(p^{3}+q)=0$
  2. $(p^{3}+q)x^{2}-(p^{3}-2q)x+(p^{3}+q)=0$
  3. $(p^{3}-q)x^{2}-(5p^{3}-2q)x+(p^{3}-q)=0$
  4. $(p^{3}-q)x^{2}-(5p^{3}+2q)x+(p^{3}-q)=0$
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B Correct answer
Explanation

Given alpha + beta = -p and alpha^3 + beta^3 = q. We know alpha^3 + beta^3 = (alpha + beta)((alpha + beta)^2 - 3alpha*beta). So q = -p(p^2 - 3alpha*beta). q = -p^3 + 3p*alpha*beta, so alpha*beta = (p^3 + q) / 3p. The roots are x1 = alpha/beta and x2 = beta/alpha. Sum = (alpha^2 + beta^2) / (alpha*beta) = ((alpha+beta)^2 - 2alpha*beta) / (alpha*beta) = (p^2 - 2alpha*beta) / (alpha*beta). Product = 1. The quadratic is x^2 - (sum)x + 1 = 0.

AI explanation

We are given alpha + beta = -p and alpha^3 + beta^3 = q. Using the sum of cubes identity, (alpha + beta)^3 - 3alpha*beta(alpha + beta) = q, we substitute -p for the sum to get -p^3 + 3p(alpha*beta) = q, which means 3p(alpha*beta) = p^3 + q and alpha*beta = (p^3 + q)/(3p). For the new equation with roots alpha/beta and beta/alpha, the sum of roots is (alpha^2 + beta^2)/(alpha*beta), which equals ((alpha + beta)^2 - 2alpha*beta)/(alpha*beta). Substituting the known values gives ((-p)^2 - 2(p^3 + q)/(3p)) / ((p^3 + q)/(3p)), simplifying to (p^3 - 2q)/(p^3 + q). The product of the new roots is (alpha/beta)(beta/alpha) = 1. Using the standard form of a quadratic equation x^2 - (sum)x + (product) = 0, we multiply by (p^3 + q) to get (p^3 + q)x^2 - (p^3 - 2q)x + (p^3 + q) = 0.

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