Solve the following equation by factorisation:

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

Given,

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

On squaring both sides, we get

$3x + 4 = x^2 \\[0.5em] \Rightarrow x^2 - 3x - 4 = 0 \\[0.5em] \Rightarrow x^2 - 4x + x - 4 = 0 \\[0.5em] \Rightarrow x(x - 4) + 1(x - 4) = 0 \\[0.5em] \Rightarrow (x + 1)(x - 4) = 0 \text{ (Factorising left side) }\\[0.5em] \Rightarrow x + 1 = 0 \text{ or } x - 4 = 0 \text{ (Zero-product rule) }\\[0.5em] x = -1 \text{ or } x = 4$

As equation is squared so roots need to be checked so putting x = -1 and x = 4 in the equation $\sqrt{3x + 4} = x.$
Checking for x = -1

$\Rightarrow \sqrt{3 \times -1 + 4} = -1 \\[0.5em] \Rightarrow \sqrt{-3 + 4} = -1 \\[0.5em] \Rightarrow \sqrt{1} = -1 \text{ (This equation is false) }$

Checking for x = 4

$\Rightarrow \sqrt{3 \times 4 + 4} = 4 \\[0.5em] \Rightarrow \sqrt{12 + 4} = 4 \\[0.5em] \Rightarrow \sqrt{16} = 4 \text{ (This equation is true ) }\\[0.5em]$

Since for x = -1 equation is false hence x = -1 is not the root of the given equation.
Hence, the root of given equation is 4.