Question

If $x=\frac{\sqrt{2p+3q}+\sqrt{2p-3q}}{\sqrt{2p+3q}-\sqrt{2p-3q}}$, find the value of $3q{x}^2-4px+3q$.

Answer 1

We have,

$x=\frac{\sqrt{2p+3q}+\sqrt{2p-3q}}{\sqrt{2p+3q}-\sqrt{2p-3q}}$

$\Rightarrow x=\frac{\sqrt{2p+3q}+\sqrt{2p-3q}}{\sqrt{2p+3q}-\sqrt{2p-3q}}\times \frac{\sqrt{2p+3q}+\sqrt{2p-3q}}{\sqrt{2p+3q}+\sqrt{2p-3q}}$

$\Rightarrow x=\frac{(\sqrt{2p+3q}+\sqrt{2p-3q}{)}^2}{(\sqrt{2p+3q}{)}^2-(\sqrt{2p-3q}{)}^2}[\because (x-y\left)\right(x+y)=x^2-y^2]$

$\Rightarrow x=\frac{2p+3q+2p-3q+2\sqrt{2p+3q}\sqrt{2p-3q}}{2p+3q-(2p-3q)}[\because (x+y{)}^2=x^2+y^2+2xy]$

$\Rightarrow x=\frac{4p+2\left(\sqrt{4p^2-9q^2}\right)}{6q}$

$\Rightarrow x=\frac{2p+\left(\sqrt{4p^2-9q^2}\right)}{3q}$

Now, put the value of $x$ in $3q{x}^2-4px+3q$, we get

$3q[\frac{2p+\left(\sqrt{4p^2-9q^2}\right)}{3q}{]}^2-4p[\frac{2p+\left(\sqrt{4p^2-9q^2}\right)}{3q}]+3q$

$=\frac{4p^2+4p^2-9q^2+4p\sqrt{4p^2-9q^2}}{3q}-\frac{8p^2+4p\sqrt{4p^2-9q^2}}{3q}+\frac{9q^2}{3q}$

$=0$