Question

If $\alpha$ 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 always

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

Answer 1

Answer: (A)

Since, $\alpha$,$\beta$ are the roots of $x^2+px+q=0$ and $\alpha^4$,$\beta^4$ are the roots of $x^2-rx+s=0.$
$\Rightarrow \alpha+\beta=-p; \alpha\beta=q; \alpha^4+\beta^4=r\,$ and $\alpha^4\beta^4=s$
Let roots of $x^2-4qx+(2q^2-r)=0 $ be $ \alpha$' and $\beta'$
Now, $\alpha'\beta'=(2q^2-r)=2(\alpha\beta)^2-(\alpha^4+\beta^4)$
$=-(\alpha^4+\beta^4-2\alpha^2\beta^2)$
$=-(\alpha^2-\beta^2)^2 < 0$
$\Rightarrow $ Roots are real and of opposite sign.