Question: Solve: 2\sqrt{x} - \sqrt{4x -3} = \dfrac{1}{\sqrt{4x-3}}

Solution:
Given,

2\sqrt{x} - \sqrt{4x -3} = \dfrac{1}{\sqrt{4x-3}}

or, 2\sqrt{x} = \dfrac{1}{\sqrt{4x-3}} + \sqrt{4x -3}

or, 2\sqrt{x} = \dfrac{1 + (\sqrt{4x-3})(\sqrt{4x-3})}{\sqrt{4x-3}}

or, 2\sqrt{x}(\sqrt{4x-3}) = 1 + \sqrt{(4x-3)(4x-3)}

or, 2\sqrt{x(4x-3)} = 1 + \sqrt{(4x-3)^2}

or, 2\sqrt{x(4x-3)} = 1+ 4x -3

or, 2\sqrt{x(4x-3)} = 4x - 2

or, 2 \sqrt{x(4x-3)} = 2(2x-1)

or, \sqrt{x(4x-3)} = 2x -1

squaring both sides

or, (\sqrt{x(4x-3)} )^2 = (2x-1)^2

or, x(4x-3) = (2x)^2 - 2 * 2x * 1 + 1^2

or, 4x^2 - 3x = 4x^2 - 4x +1

or, 4x^2 - 4x^2 -3x +4x = 1

or, 0 + x = 1

\therefore x = 1
= Answer