Question: Solve: \sqrt{4x -3} + \sqrt{2x +3} = 6

Solution:
Given,

\sqrt{4x -3} + \sqrt{2x +3} = 6

or, \sqrt{4x -3} = 6- \sqrt{2x +3}

Squaring both sides

or, ( \sqrt{4x -3})² = (6- \sqrt{2x +3})²

or, 4x -3 = 6² - 2×6×\sqrt{2x +3} + (\sqrt{2x +3})²

or, 4x -3 = 36 - 12\sqrt{2x +3} +2x +3

or, 4x -3 = 39 + 2x -12\sqrt{2x +3}

or, 4x -2x -3 -39 = -12\sqrt{2x +3}

or, 2x -42 = 12\sqrt[2x +3}

or, 2(x -21) = 2×6 \sqrt{2x+3}

or, x -21 = 6\sqrt{2x +3}

Squaring both sides

or, (x-21)² = (6 \sqrt{2x +3})²

or, x² - 2×x×21 + 21² = 36(2x +3)

or, x² - 42x +441 = 72x +108

or, x² -42x -72x +441 -108 = 0

or, x² - 114x +333 = 0

or, x² -(111+3)x +333 = 0

or, x² -111x -3x +333 = 0

or, x(x-111) -3(x -111) = 0

or, (x-3)(x-111) = 0

Either,

(x -3) = 0

\therefore x = 3

Or,

(x-111) = 0

\therefore x = 111

Now,

Substituting the value of x = 3 in given equation:

\sqrt{4×3 -3} + \sqrt{2×3 +3} = 6

or, \sqrt{9} + \sqrt{9} = 6

or, 3 + 3 = 6

or, 6 = 6 which is true.

Again,

Substituting the value of x= 111 in given equation:

\sqrt{4×111 - 3} + \sqrt{2×111 +3} = 6

or, \sqrt{441} + \sqrt{225} = 6

or, 21 + 15 = 6

or, 36 = 6 which is false.

Hence, the value of x = 3.