Question

If $x=\frac{\sqrt{p+q}+\sqrt{p-q}}{\sqrt{p+q}-\sqrt{p-q}}$ find value of $q{x}^2-2px+q$

Answer 1

Property: If $\frac{a}{b}=\frac{c}{d}$ then,

$\frac{a+b}{a-b}=\frac{c+d}{c-d}$

Now,

$\frac{x}{1}=\frac{\sqrt{p+q}+\sqrt{p-q}}{\sqrt{p+q}-\sqrt{p-q}}$

Using the property,

$\frac{x+1}{x-1}=\frac{\sqrt{p+q}+\sqrt{p-q}+\sqrt{p+q}-\sqrt{p-q}}{\sqrt{p+q}+\sqrt{p-q}-\sqrt{p+q}+\sqrt{p-q}}$

$\frac{x+1}{x-1}=\frac{\sqrt{p+q}}{\sqrt{p-q}}$

Squaring both sides,

${\left(\frac{x+1}{x-1}\right)}^2=\frac{p+q}{p-q}$

$\frac{(x^2+1)+2x}{(x^2+1)-2x}=\frac{p+q}{p-q}$

$\Rightarrow \frac{x^2+1}{2x}=\frac{p}{q}$

$(x^2+1)q-2px=0$

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