Question

If the roots of the equation $x^2+px+q=0$, and ${\alpha }^4$ and ${\beta }^4$ are the roots of $x^2-rx+q=0$,
the roots of $x^2-4qx+2q^2-r=0$ are always

• A. Both non-real
• B. Both positive
• C. Both opposite
• D. Opposite in sign

Answer 1

Answer: (D)

Opposite in sign

Since, α,β are the roots of $x_2+px+q=0$ and ${\alpha }^4,{\beta }^4$ are the roots of $x_2-rx+s=0$.

$⟹$α+β=−p;

$⟹$αβ=q

;$⟹{\alpha }^4+{\beta }^4=r$ and ${\alpha }^4{\beta }^4=s$

Let roots of $x^2-4qx+(2q^2-r)=0$ be α′and β′

Now, α′β′=$(2q^2-r)=2(\alpha \beta {)}^2-({\alpha }^4+{\beta }^4)$

$=-({\alpha }^4+{\beta }^4-2{\alpha }^2{\beta }^2)$

$=-({\alpha }^2-{\beta }^2{)}^2<0$

$⟹$ Roots are real and of opposite sign.