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

Answer 1

Answer: (A), (D)

both real

Let $\alpha$ and $\beta$ be the roots of $x^2+ax+b=0.$

$\Rightarrow {\alpha }^4$ and ${\beta }^4$ be the roots of $x^2-cx+d=0.$

$\alpha +\beta =-a,\alpha \beta =b$ and ${\alpha }^4+{\beta }^4=c,(\alpha \beta {)}^4=d$
$\Rightarrow b^4=d$

${\alpha }^4+{\beta }^4=c$
$\Rightarrow ({\alpha }^2+{\beta }^2{)}^2-2(\alpha \beta {)}^2=c$
$\Rightarrow \left(\right(\alpha +\beta {)}^2-2\alpha \beta {)}^2-2(\alpha \beta {)}^2=c$
$\Rightarrow \left(\right(a{)}^2-2b{)}^2-2(b{)}^2=c$
$\Rightarrow (a^2-2b{)}^2-2b^2=c\Rightarrow 2b^2+c=(a^2-2b{)}^2...\left(1\right)\Rightarrow 2b^2+c=a^4+4b^2-4a^{2}b\Rightarrow 2b^2-c=a^2(4b-a^2)...\left(2\right)$

Now, for equation
$x^2-4bx+2b^2-c=0$
$D=(4b{)}^2–4\cdot 1\cdot (2b^2–c)$
$=4(2b^2+c)$
$=4(a^2–2b{)}^2$ [from eqn(1)]
$\Rightarrow D>0$
Hence, roots are real.

Now, we have $f\left(x\right)=x^2-4bx+2b^2-c$ $f\left(0\right)=2b^2-c$
From eqn(2),
$2b^2-c=a^2(4b-a^2)<0(\because a^2>4b)$
$\Rightarrow$ Roots are opposite in sign.