Question

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

• A. $x^2+5x+1=0$
• B. $x^2-x+1=0$
• C. $x^2+x-1=0$
• D. $x^2+x+1=0$

Answer 1

The correct option is D $x^2+x+1=0$

$x^3+125=0$

$\Rightarrow {\left(\frac{x}{-5}\right)}^3=1$

$\therefore x=-5,-5\omega ,-5{\omega }^2$

Let $\alpha =-5,\beta =-5\omega ,\gamma =-5{\omega }^2$

$S={\left(\frac{\alpha }{\beta }\right)}^2+{\left(\frac{\alpha }{\gamma }\right)}^2=\omega +{\omega }^2$

and

product of roots $\omega .{\omega }^2=1$

$\therefore$ Required quadratic equation

$x^2-(\omega +{\omega }^2)x+{\omega }^2.\omega =0$

$\Rightarrow x^2+x+1=0$