Multiple choice

If $ \alpha + \beta = 3$ and $ \alpha^3 + \beta^3 = 9$, find the quadratic equation whose roots are $ \alpha$ or $\beta$ :

  1. $x^2-3x+3=0$
  2. $x + \cfrac{2}{x}+ 3 = 0$
  3. $x^2- 2x + 3 = 0$
  4. $x + \cfrac{2}{x} = 3$
  5. $x^2-3x+2=0$
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E Correct answer
Explanation

Given alpha + beta = 3 and alpha^3 + beta^3 = 9. Using (alpha + beta)^3 = alpha^3 + beta^3 + 3*alpha*beta*(alpha + beta), we get 3^3 = 9 + 3*alpha*beta*(3), so 27 = 9 + 9*alpha*beta, meaning alpha*beta = 2. The quadratic equation is x^2 - (sum)x + (product) = 0, which is x^2 - 3x + 2 = 0.

AI explanation

Using the identity for the sum of cubes, alpha^3 + beta^3 = (alpha + beta)^3 - 3*alpha*beta*(alpha + beta). Substituting the given values 3 and 9, we get 9 = 3^3 - 3*alpha*beta*3, which simplifies to 9 = 27 - 9*alpha*beta, yielding a product (alpha * beta) of 2. Using Vieta's formulas, the required quadratic equation is x^2 - (sum)x + (product) = 0. Therefore, the equation is x^2 - 3x + 2 = 0.

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