Question

lf $\alpha$ and $\beta$ are the roots of $x^2+px+q=0$ and ${\alpha }^4,{\beta }^4$ are the roots of $x^2-rx+s=0$, then the equation $x^2-4qx+(2q^2-r)=0$ has

• A. two real roots
• B. two negative roots
• C. two positive roots
• D. one positive and one negative root

Answer 1

Answer: (D)

one positive and one negative root

$f\left(0\right)=2q^2-r=2(\alpha \beta {)}^2-({\alpha }^4+{\beta }^4)$
$=-({\alpha }^2-{\beta }^2{)}^2<0.$
$\therefore f\left(0\right)<0$

And co-efficient of $x^2$ is greater than zero.
$\Rightarrow$ It has one positive and one negative root.