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 ${\beta }^{\prime }$
Now, ${\alpha }^{\prime }{\beta }^{\prime }=(2q^2-r)=2(\alpha \beta {)}^2-({\alpha }^4+{\beta }^4)$
$=-({\alpha }^4+{\beta }^4-2{\alpha }^2{\beta }^2)$
$=-({\alpha }^2-{\beta }^2{)}^2<0$
$\Rightarrow$ Roots are real and of opposite sign.