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

$or, (\sqrt{x(5+x)} )² = (10-x)²$

$or, x(5+x) = 10² -2×10×x + x²$

$or, 5x +x² = 100 - 20x +x²$

$or, x² - x² +5x +20x = 100$

$or, 25x = 100$

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

$\therefore x = 4$