Question

If $\alpha$ and $\beta$ are the roots of $x^2+px+q=0$, then the value of $(\omega \alpha +{\omega }^2\beta )({\omega }^2\alpha +\omega \beta )$ where $\omega$ is cube root of unity is

• A. $p^2-3q$
• B. $p^2-2q$
• C. $p-2q$
• D. $2p-q$

Answer 1

The correct option is A $p^2-3q$

Given, quadratic equation: $x^2+px+q=0$ and its root $=\alpha$ and $\beta$
We know that the standard quadratic equation is: $a{x}^2+bx+c=0$
Comparing the given equation with the standard equation, we get $a=1,b=p$ and $c=q$.
We also know that sum of the roots $(\alpha +\beta )=-\frac{b}{a}=-\frac{p}{1}=-p$.
And product of the roots $\left(\alpha \beta \right)=\frac{c}{a}=\frac{q}{1}=q$.
Also, $1+\omega +{\omega }^2=0$ or $\omega +{\omega }^2=-1$ and ${\omega }^3=1$
Therefore, $(\omega \alpha +{\omega }^2\beta )({\omega }^2\alpha +\omega \beta )={\omega }^3\left({\alpha }^2{\beta }^2\right)+{\omega }^2\alpha \beta +{\omega }^4\alpha \beta ={\alpha }^2+{\beta }^2+\alpha \beta ({\omega }^2+\omega )$$={\alpha }^2+{\beta }^2-\alpha \beta =(\alpha +\beta {)}^2-3\alpha \beta =p^2-3q$.