Question

If $x=\frac{\sqrt{p+2q}+\sqrt{p-2q}}{\sqrt{p+2q}-\sqrt{p-2q}}$, show that $q{x}^2-px+q=0$

Answer 1

$x=\frac{\sqrt{p+2q}+\sqrt{p-2q}}{\sqrt{p+2q}-\sqrt{p-2q}}$

$x=\frac{\sqrt{p+2q}+\sqrt{p-2q}}{\sqrt{p+2q}-\sqrt{p-2q}}\times \frac{\sqrt{p+2q}+\sqrt{p-2q}}{\sqrt{p+2q}+\sqrt{p-2q}}$

$⟹x=\frac{p+2q+p-2q+2\sqrt{p^2-4q^2}}{p+2q-p+2q}$

$⟹x=\frac{p+\sqrt{p^2-4q^2}}{2q}$

Comparing with root $\frac{-b+\sqrt{b^2-4ac}}{2a}$ of equation $a{x}^2+bx+c=0$ which satisfy this quadratic equation.

For calculated $x$, $b=-p,a=q,c=c=q$

Hence given value of x satisfy equation

$q{x}^2-px+q=0$