Question: Solve: $\sqrt{3x +1} - \sqrt{x -1} = 2$

Solution:
Given,

$\sqrt{3x +1} - \sqrt{x -1} = 2$

$or, \sqrt{3x +1} = 2 + \sqrt{x -1}$

[ Squaring both sides ]

$or, (\sqrt{3x +1})² = (2+\sqrt{x -1})²$

$or, 3x +1 = 2² + 2×2×\sqrt{x -1} + (\sqrt{x-1})²$

$or, 3x +1 = 4 + 4\sqrt{x -1} + x -1$

$or, 3x -x = 4-1 -1 + 4\sqrt{x -1}$

$or, 2x = 2 + 4\sqrt{x -1}$

$or, 4\sqrt{x -1} = 2x -2$

$or, 4\sqrt{x -1} = 2(x-2)$

$or, 2\sqrt{x-1} = (x-1)$

$or, 2 = \dffac{x-1}{\sqrt{x}-1}$

$or, 2 = \dfrac{(sqrt{x})² - (1)²}{\sqrt{x} -1}$

$or, 2 = \dfrac{(\sqrt{x} +1)(\sqrt{x} -1)}{(\sqrt{x} -1)}$

$or, 2 = \sqrt{x} +1$

$or, 2-1 = \sqrt{x}$

$or, \sqrt{x} = 1$

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

$\therefore x = 1$
= Answer