Nature of roots of the equation $x^2-4qx+2q^2-r=0$

Question

If $\alpha$ and $\beta$ are roots of $x^2+px+q=0$ and $(\alpha)^4$ and $(\beta)^4$ are the roots of the equation $x^2-rx+s$ then the equation $x^2-4qx+2q^2-r=0$ has always

(A)two real roots (B)two positive roots (C)two negative roots (D)one positive and one negative root

Source:JEE 1990

Ok this one seems quite tricky.I have found sum of roots and product of roots to the first two equations.But how can I incorporate them to find the nature of root of the third?

Answer 1

$r=\alpha^4+\beta^4=(p^2-2q)^2-2q^2=p^4-4p^2q+2q^2 $ . So

$2q^2-r=(4q-p^2)p^2=-D_{1}p^2$ where $D_1$ is the deicriminant of the first equation $D_1=p^2-4q$

Now $D_3=16q^2-4(2q^2-r)=8q^2+4r=4(p^2-2q)^2$ . So equation will have two real roots. Now find the exact roots of the third equation by quadratioc formulae -:

$l=\frac{4q-2(p^2-2q)}{2}=4q-p^2$ & $m=\frac{4q+2(p^2-2q)}{2}=p^2$

Now if $\alpha,\beta$ are real then $p^2-4q>0$ which means third equation has one positive and one negative root. Otherwise it has both positive roots.

So $A,B$ & $D$ are correct options

Hope this will be helpful !

Answer 2

$$r=\alpha^4+\beta^4=(\alpha^2+\beta^2)^2-2q^2=(p^2-2q)^2-2q^2=p^4-4p^2q^2+2q^2$$

Now the discriminant of the last equation is $$(4q)^2-4(2q^2-r)=16q^2-4(4p^2q^2-p^4)=(4q^2-p^2)^2\ge0$$