Multiple choice

If $\alpha ,\beta ,\gamma $ are roots of the equation ${ x }^{ 3 }+125=0$, then the quadratic equation whose roots are ${ \left( \cfrac { \alpha }{ \beta } \right) }^{ 2 }$ and ${ \left( \cfrac { \alpha }{ \gamma } \right) }^{ 2 }$ is

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

The roots of x^3 + 125 = 0 are -5, -5w, -5w^2 where w is a complex cube root of unity. The ratios alpha/beta and alpha/gamma are w and w^2 (or similar permutations). Squaring these gives w^2 and w^4 = w. The quadratic equation with roots w and w^2 is x^2 - (w+w^2)x + w*w^2 = 0, which simplifies to x^2 + x + 1 = 0.

AI explanation

The roots of x^3 + 125 = 0 are found by taking the cube roots of -5, which are -5, and the complex cube roots of unity scaled by 5. Let the roots be alpha = -5, beta = 5w, and gamma = 5w^2, where w is a complex cube root of unity. The required roots are (alpha/beta)^2 and (alpha/gamma)^2, which simplify to 1/w^2 and 1/w. Using the properties of cube roots of unity, these values are w and w^2. The sum of these roots is w + w^2 = -1 and the product is w * w^2 = w^3 = 1. Applying Vieta's formulas, the required quadratic equation is x^2 - (sum)x + product = 0, resulting in x^2 + x + 1 = 0.

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