Question

If $\alpha ,\beta$ are the real and distinct roots of $x^2+px+q=0$ and ${\alpha }^4,{\beta }^4$ are the roots of $x^2-rx+s=0$, then the equation $x^2-4qx+2q^2-r=0$ has always.

• A. Two real roots
• B. Two negative roots
• C. Two positive roots
• D. One positive root and one negative root

Answer 1

The correct option is A Two real roots

Given, Reacts of $x^2+px+q=0$

Roots $=a,p$

and root of $x^2-rx+s=0$

Roots $=x^2,p^4$

so, $\alpha +\beta =-p$ and $\alpha \beta =q$ (equation $\left(1\right)$)

and, ${\alpha }^2+{\beta }^4=({\alpha }^2+{\beta }^2{)}^2=2{\alpha }^2{\beta }^2$

$=\left(\right(\alpha +\beta {)}^2-2\alpha \beta {)}^2-2{\alpha }^2{\beta }^2$

so, $r=(p^2-2q{)}^2-2(q{)}^2$

$=p^2-4p^{2}q+4p^2-2q^2$

$=p^4-4p^{2}q+2q^2$

$\Rightarrow 2q^2-r=p^2(4q-p^2)<0$

as $p^2>0$ or $4q-p^2<0$ (from $\left(1\right)$)

$\Rightarrow x^2-4px+2q^2-r=0$

for the equation have real & distance roots $0>0\theta =16p^2-4(2q^2-r)$

$=16p^2-4$

so, $\theta >0$