Question: Solve: $\sqrt{x +8} -2 = \sqrt{x}$

Solution:
Given,

$\sqrt{x +8} -2 = \sqrt{x}$

$or, \sqrt{x+8} = \sqrt{x} + 2$

[ Squaring both sides ]

$or, (\sqrt{x+8})^2 = (\sqrt{x} + 2)^2$

$or, x +8 = (\sqrt{x})^2 + 2×\sqrt{x}×2 + 2^2$

$or, x + 8 = x + 4\sqrt{x} + 4$

$or, x -x + 8 -4 = 4\sqrt{x}$

$or, 4 = 4\sqrt{x}$

$or, 4\sqrt{x} = 4$

$or, \sqrt{x} = \dfrac{4}{4}$

$or, \sqrt{x} = 1$

$or, \sqrt{x} = \sqrt{1}$

$\therefore x = 1$
= Answer