If $\dfrac{\sqrt{1+x} + \sqrt{1-x}}{\sqrt{1+x} - \sqrt{1-x}} = p$, then $p^2 - \dfrac{2p}{x} + 1 =$

Rationalizing gives $\frac{p+1}{p-1}=\sqrt{\frac{1+x}{1-x}}$; squaring and applying componendo–dividendo yields $x=\frac{2p}{p^2+1}$, i.e. $p^2+1=\frac{2p}{x}$. Rearranging gives $p^2-\frac{2p}{x}+1=0$.