Solve the following equation by factorisation:

$\sqrt{x(x - 7)} = 3\sqrt{2}$

Given,

$\sqrt{x(x - 7)} = 3\sqrt{2}$

On squaring both sides, we get

$x(x - 7) = 18 \\[0.5em] \Rightarrow x^2 - 7x = 18 \\[0.5em] \Rightarrow x^2 - 7x - 18 = 0 \\[0.5em] \Rightarrow x^2 - 9x + 2x - 18 = 0 \\[0.5em] \Rightarrow x(x - 9) + 2(x - 9) = 0 \\[0.5em] \Rightarrow (x + 2)(x - 9) = 0 \text{ (Factorising left side) } \\[0.5em] \Rightarrow x + 2 = 0 \text{ or } x - 9 = 0 \text{ (Zero-product rule) } \\[0.5em] x = -2 \text{ or } x = 9.$

As equation is squared so roots need to be checked so putting x = -2 and x = 9 in the equation $\sqrt{x(x - 7)} = 3\sqrt{2}$
Checking for x = -2

$\Rightarrow \sqrt{-2(-2 - 7)} = 3\sqrt{2} \\[0.5em] \Rightarrow \sqrt{-2(-9)} = 3\sqrt{(2)}\\[0.5em] \Rightarrow \sqrt{18} = 3\sqrt{2} \text{ (This equation is true) }\\[0.5em]$

Checking for x = 9

$\Rightarrow \sqrt{9(9 - 7)} = 3\sqrt{2} \\[0.5em] \Rightarrow \sqrt{9 \times 2} = 3\sqrt{2} \\[0.5em] \Rightarrow \sqrt{18} = 3\sqrt{2} \text{ (This equation is true) }$

As the above two equations are true,

∴ The roots of given equation are -2, 9.