Question

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

Answer 1

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

Rationalizing

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

$=\frac{[\sqrt{p+2q}+\sqrt{p-2q}{]}^2}{(p+2q)-(p-2q)}$

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

Putting value of $x$ in $q{x}^2-px+q$

$q\frac{[\sqrt{p+2q}+\sqrt{p-2q}{]}^2}{4q}-p\frac{[\sqrt{p+2q}+\sqrt{p-2q}{]}^2}{4q}+q$

$=\frac{q[p+2q+p-2q+2\sqrt{p+2q}\sqrt{p-2q}]}{4q}-\frac{p[p+2q+p-2q+2\sqrt{p+2q}\sqrt{p-2q}]}{4q}+q$

$=\frac{q[p+\sqrt{p^2-2q^2}{]}^2}{2q}-p\frac{[2p+2\sqrt{p^2-2q^2}]}{4q}+q$

$=\frac{q[p+\sqrt{p^2-2q^2}{]}^2}{2q}-\frac{p[p+\sqrt{p^2-2q^2}]}{2q}+q$

$=p^2+2q^2-2q^2+2p\sqrt{p^2-2q^2}-2p^2+2p\sqrt{p^2-2q^2}/2q+2q$

$=-2q^2/2q+2q$

$=-2q+2q=0$