Question

If the roots of the equation $x^2+px+q=0$ are $\alpha$ and $\beta$ and roots of the equation $x^2-xr+s=0$ are ${\alpha }^4,{\beta }^4$, then the roots of the equation $x^2-4qx+2q^2-r=0$ will be

• A. Both negative
• B. Both positive
• C. Both real
• D. One negative and one positive

Answer 1

The correct option is C Both real

Given, $\alpha ,\&\beta$ be the root.

We know that $\alpha +\beta =-p,\alpha .\beta =q$

$\Rightarrow {\alpha }^4+{\beta }^4=r,{\alpha }^4.{\beta }^4=s$

$\Rightarrow {\alpha }^2+{\beta }^2=(\alpha +\beta {)}^2-2\alpha .\beta =p^2-2q$

To find ${\alpha }^4+{\beta }^4$

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

$\Rightarrow r=(p^2-2q{)}^2-2q^2=p^4+4q^2-4p^{2}q-2q^2$

$\Rightarrow 3(p{)}^2-4q{p}^2+2q^2-r=0$

$\Rightarrow x=p^2$ satisfy root

other root $\Rightarrow \nu +p^2=4q\Rightarrow \nu =4q-p^2$

$\Rightarrow r{p}^2=2q^2-r$

$\Rightarrow 0=16q^2-4(2q^2-r)\Rightarrow 8q^2+4r\Rightarrow 8q^2+4\left(\right(p^2-2q{)}^2-2q^2)\Rightarrow 4(p^2-2q{)}^2\ge 0p^2,-p^2+4q\therefore \text{Both are real value}$