Question

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

Answer 1

Given:-

$\alpha$ and $\beta$ are the roots of equation $x^2+px+q=0$

${\alpha }^4$ and ${\beta }^4$ re the roots of equation $x^2-rx+s=0$

To Prove:- Equation $x^2-4qx+2q^2-r=0$ has real roots.

Proof:-

$\alpha$ and $\beta$ are the roots of equation $x^2+px+q=0$

Therefore,

$q=\alpha \beta .....\left(1\right)(\because \text{Product of roots}=\frac{c}{a})$

Again,

${\alpha }^4$ and ${\beta }^4$ re the roots of equation $x^2-rx+s=0$

$r={\alpha }^4+{\beta }^4.....\left(2\right)(\because \text{sum of roots}=\frac{-b}{a})$

Now,

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

For roots to be real,

$D>0$

Therefore,

$b^2-4ac$

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

$=16q^2-8q^2+4r$

$=8q^2+4r$

$=4(2q^2+r)$

$=4[2{\left(\alpha \beta \right)}^2+({\alpha }^4+{\beta }^4)]\left(\text{From}\left(1\right)\&\left(2\right)\right)$

$=4[2{\alpha }^2{\beta }^2+({\alpha }^4+{\beta }^4)]$

$\Rightarrow =4\left[{({\alpha }^2+{\beta }^2)}^2\right]>0$

Therefore the equation $x^2-4qx+2q^2-r=0$ has real roots.

Hence proved.