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