Question: Solve: $\sqrt{x} + \sqrt{x+13} = \dfrac{91}{\sqrt{x+13}}$

Solution:
Given,

 $\sqrt{x} + \sqrt{x+13} = \dfrac{91}{\sqrt{x+13}}$

$or, \sqrt{x}  = \dfrac{91}{\sqrt{x+13}} - \sqrt{x+13}$

$or, \sqrt{x} = \dfrac{91 - (\sqrt{x+13})(\sqrt{x+13})}{\sqrt{x+13}}$

$or, \sqrt{x} (\sqrt{x + 13} = 91 - \sqrt{(x+13)^2}$

$or, \sqrt{x(x+13)} = 91 - (x+13)$

$or, \sqrt{x(x+13)} = 91 -x -13$

$or, \sqrt{x(x+13)} = 78 - x $

Squaring both sides

$or, (\sqrt{x(x+13)})^2 = (78-x)^2 $

$or, x(x+13) = 78^2 - 2*78*x + x^2 $

$or, x^2 + 13x = 6084 - 156 x + x^2$

$or, x^2 - x^2 +13x + 156c = 6084$

$or, 169x = 6084$

$or, x = \dfrac{6084}{169}$

$\therefore x = 36$
= Answer