Solve for $x$ when $\;\;$ $\dfrac{\sqrt{a + x} + \sqrt{a - x}}{\sqrt{a + x}- \sqrt{a - x}} = b$
Given: $\;$ $\dfrac{\sqrt{a + x} + \sqrt{a - x}}{\sqrt{a + x}- \sqrt{a - x}} = b = \dfrac{b}{1}$
By componendo-dividendo we have,
$\dfrac{\left(\sqrt{a + x} + \sqrt{a - x}\right) + \left(\sqrt{a + x} - \sqrt{a - x}\right)}{\left(\sqrt{a + x} + \sqrt{a - x}\right) - \left(\sqrt{a + x} - \sqrt{a - x}\right)} = \dfrac{b + 1}{b - 1}$
i.e. $\;$ $\dfrac{2 \sqrt{a + x}}{2 \sqrt{a - x}} = \dfrac{b + 1}{b - 1}$
i.e. $\;$ $\dfrac{\sqrt{a + x}}{\sqrt{a - x}} = \dfrac{b + 1}{b - 1}$
Squaring both sides we get,
$\dfrac{a + x}{a - x} = \dfrac{\left(b + 1\right)^2}{\left(b - 1\right)^2}$
i.e. $\;$ $\dfrac{a + x}{a - x} = \dfrac{b^2 + 2b + 1}{b^2 - 2b + 1}$
By componendo-dividendo we have,
$\dfrac{\left(a + x\right) + \left(a - x\right)}{\left(a + x\right) - \left(a - x\right)} = \dfrac{\left(b^2 + 2b + 1\right) + \left(b^2 - 2b + 1\right)}{\left(b^2 + 2b + 1\right)- \left(b^2 - 2b + 1\right)}$
i.e. $\;$ $\dfrac{2a}{2x} = \dfrac{2 \left(b^2 + 1\right)}{4b}$
i.e. $\;$ $\dfrac{a}{x} = \dfrac{b^2 + 1}{2b}$
$\implies$ $x = \dfrac{2ab}{b^2 + 1}$