√{3x +4} +x = 12 | Surds - Simplify | SciPiPupil

Question: Solve: $\sqrt{3x +4} + x = 12$

Solution:

Given,

$\sqrt{3x +4} + x = 12$

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

[ Squaring both sides ]

$or, (\sqrt{3x +4})^2 = (12-x)^2$

$or, 3x +4 = 12^2 - 2×12×x + x^2$

$or, 3x +4 = 144 - 24x + x^2$

$or, x^2 - 24x -3x +144-4 = 0$

$or, x^2 - 27x +140 = 0$

$or, x^2 - (20+7)x +140 = 0$

$or, x^2 - 20x -7x + 140 = 0$

$or, x(x -20) -7(x -20) = 0$

$or, (x -7)(x -20) = 0$

Either,

$(x -7) = 0$

$\therefore x = 7$

Or,

$(x -20) = 0$

$\therefore x = 20$

Now,

Substituting x=7 in the given equation,

$\sqrt{3×7 +4} +7 = 12$

$or, \sqrt{25} + 7 = 12$

$or, 5 + 7 = 12$

$or, 12 = 12$ which is true.

Again,

Substituting x = 20 in the given equation,

$\sqrt{3×20 +4} + 7= 12$

$or, \sqrt{64} + 7 = 12$

$or, 8 + 7 = 12$

$or, 15 = 12$ which is false.

Hence, x = 7.