Question

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

Answer 1

The equation $x^2-4qx+2q^2-r=0$

$D=b^2-4ac$

$\therefore D=(-42{)}^2-4(1\left)\right(2q^2-r)$

$=16q^2-8q^2+4r$ $=\sqrt{8q^2}+4r$

Now ${\alpha }^4+{\beta }^4=\frac{-(-r)}{1}=r$

$\therefore {\alpha }^4+{\beta }^4>0$ $({\alpha }_1\beta \ne 0)$

$\therefore r>0$

$\therefore 8q^2+4r>0$ $(D>0)$ proved

$\therefore$ The equation $x^2-4qx+2q^2-r=0$ has real and distinct roots.