Multiple choice

Choose the correct answer from the alternatives given. If a and $\beta$ be the roots of the equation $ax^2 \, + \, bx \, + \, c \, = \, 0,$ then find the values of $\dfrac{\alpha^2}{\beta} \, + \, \dfrac{\beta^2}{\alpha}.$

  1. $\displaystyle \frac{ab \, - \, b^2 \, c}{2b^2 \, c}.$
  2. $\displaystyle \frac{3bc \, - \, a^2}{2b^2 \, c}.$
  3. $\displaystyle \frac{3ac \, - \, b^2}{a^3 \, c}.$
  4. $\displaystyle \frac{3abc \, - \, b^3}{a^2 \, c}.$
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D Correct answer
Explanation

alpha^2/beta + beta^2/alpha = (alpha^3 + beta^3)/(alpha*beta). alpha+beta = -b/a, alpha*beta = c/a. alpha^3+beta^3 = (alpha+beta)^3 - 3*alpha*beta*(alpha+beta) = (-b/a)^3 - 3*(c/a)*(-b/a) = -b^3/a^3 + 3bc/a^2 = (3abc - b^3)/a^3. Dividing by c/a gives (3abc - b^3)/(a^2*c).

AI explanation

By Vieta's formulas for the equation ax^2 + bx + c = 0, the sum of the roots is alpha + beta = -b/a and the product is alpha * beta = c/a. The expression alpha^2/beta + beta^2/alpha is combined into a single fraction as (alpha^3 + beta^3) / (alpha * beta). Using the identity for the sum of cubes, the numerator becomes (alpha + beta)^3 - 3 * alpha * beta * (alpha + beta). Substituting the Vieta's formulas yields ((-b/a)^3 - 3(c/a)(-b/a)) / (c/a), which simplifies to (3abc - b^3) / (a^2 * c).

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