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 roos 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)

The correct options are
A both real
D one positive and one negative

Let $a^2+ax+b=0$ has roots $\alpha$ and $\beta$
$x^2-cx+d=0$ roots are ${\alpha }^4$ and ${\beta }^4$
$\alpha +\beta =-a,\alpha \beta =b$ and ${\alpha }^4+{\beta }^4=c,{\left(\alpha \beta \right)}^4=d$
$\Rightarrow b^4=d$ and ${\alpha }^4+{\beta }^4=c$
${({\alpha }^2+{\beta }^2)}^2-2{\left(\alpha \beta \right)}^2=c$
${({(\alpha +\beta )}^2-2\alpha )}^2-2{\left(\alpha \beta \right)}^2=c$
${(a^2-2b)}^2-2b^2=c\Rightarrow 2b^2+c={(a^2-2b)}^2$
$2b^2-c=4a^{2}b-a^2=a^2(4b-a^2)$
Now for equation
$x^2-4bx+2b^2-c=0$
$D={\left(4b\right)}^2-4\left(1\right)(2b^2-c)=16b^2-8b^2+4c=8b^2+4c=4(2b^2+c)=4{(a^2-2b)}^2>0\Rightarrow$ real roots
Now $f\left(0\right)=2b^2-c=a^2(4b-a^2)<0$
Roots are opposite sign