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