Multiple choice

A quadratic equation whose roots are $\displaystyle{\left(\frac{\gamma}{\alpha}\right)^2}$ and $\displaystyle{\left(\frac{\beta}{\alpha}\right)^2}$, where $\alpha, \beta, \gamma$ are the roots of $x^3 + 27 = 0$ is

  1. $x^2 - x + 1 = 0$
  2. $x^2 + 3x + 9 = 0$
  3. $x^2 + x + 1 = 0$
  4. $x^2 - 3x + 9 = 0$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

The roots of x^3 + 27 = 0 are -3, -3w, and -3w^2, where w is a complex cube root of unity. Let alpha = -3, beta = -3w, gamma = -3w^2. The required roots are (gamma/alpha)^2 = (w^2)^2 = w^4 = w and (beta/alpha)^2 = (w)^2 = w^2. The quadratic equation with roots w and w^2 is x^2 - (w+w^2)x + w^3 = 0, which simplifies to x^2 + x + 1 = 0.

AI explanation

The roots of x^3 + 27 = 0 are found by solving x^3 = -27, giving one real root alpha = -3 and two complex roots beta = (3 + 3sqrt(3)i)/2 and gamma = (3 - 3sqrt(3)i)/2. The new equation requires roots of (gamma/alpha)^2 and (beta/alpha)^2, which evaluate to (1 - sqrt(3)i)^2 / 4 and (1 + sqrt(3)i)^2 / 4, yielding (-1 - sqrt(3)i)/2 and (-1 + sqrt(3)i)/2. For a quadratic equation x^2 - Sx + P = 0, the sum S is -1 and the product P is 1. Substituting these values gives the required quadratic equation x^2 + x + 1 = 0.

Keep practicing — related questions