Question

If the equation $x^2-cx+d=0$ has roots equal to the fourth powers of the roots of $x^2+ax+b=0$, where $a^2>4b$, then the roots of $x^2-4bx+2b^2-c=0$ will be

• A. both real
• B. both negative
• C. both positive
• D. one positive and one negative

Answer 1

Answer: (A), (D)

Let $\alpha$ and $\beta$ are the roots of $x^2+ax+b=0$
and the roots of $x^2-cx+d=0$ are ${\alpha }^4$ and ${\beta }^4$.
Now, $\alpha +\beta =-a,\alpha \beta =b\dots$(i)
and ${\alpha }^4+{\beta }^4=c,{\alpha }^4{\beta }^4=d\dots$(ii)
From Eqs. (i) and (ii),
$b^4=d$ and ${\alpha }^4+{\beta }^4=c$
${\left({\alpha }^2+{\beta }^2\right)}^2-2(\alpha \beta {)}^2=c$
$\Rightarrow {\left[(\alpha +\beta {)}^2-2\alpha \beta \right]}^2-2(\alpha \beta {)}^2=c$
$\Rightarrow {\left(a^2-2b\right)}^2-2b^2=c$
$\Rightarrow 2b^2+c={\left(a^2-2b\right)}^2$
$\Rightarrow 2b^2+c=a^4+4b^2-4a^{2}b$
$\Rightarrow 2b^2-c=4a^{2}b-a^4$
$\Rightarrow 2b^2-c=a^2\left(4b-a^2\right)$
and for equation $x^2-4bx+2b^2-c=0$
$D=(4b{)}^2-4(1)\left(2b^2-c\right)$
$=16b^2-8b^2+4c=8b^2+4c$
$=4\left(2b^2+c\right)=4{\left(a^2-2b\right)}^2>0=$ real root
Now, $f(0)=2b^2-c=a^2\left(4b-a^2\right)<0$
$\left(\because a^2>4b\right)$
$=$ roots are opposite in sign