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