Question: Solve: \sqrt{x} + \sqrt{5+x} = \dfrac{15}{\sqrt{5+x}}

Solution:
Given,

\sqrt{x} + \sqrt{5+x} = \dfrac{15}{\sqrt{5+x}}

or, \sqrt{x} = \dfrac{15}{\sqrt{5+x} }- \sqrt{5+x}

or, \sqrt{x} = \dfrac{15 - (\sqrt{5+x})(\sqrt{5+x})}{\sqrt{5+x}}

or, \sqrt{x} = \dfrac{15 - \sqrt{(5+x)(5+x)}}{\sqrt{5+x}}

or, \sqrt{x} (\sqrt{5+x}) = 15 - \sqrt{(5+x)²}

or, \sqrt{x(5+x)} = 15 - (5+x)

or, \sqrt{x(5+x)} = 15 - 5-x

or, \sqrt{x(5+x)} = 10 - x

Squaring both sides

or, (\sqrt{x(5+x)} )² = (10-x)²

or, x(5+x) = 10² -2×10×x + x²

or, 5x +x² = 100 - 20x +x²

or, x² - x² +5x +20x = 100

or, 25x = 100

or, x = \dfrac{100}{25}

\therefore x = 4