# How do you solve  sqrt(12-n)=n and find any extraneous solutions?

Aug 28, 2016

$n = 3$

#### Explanation:

Given:

$\sqrt{12 - n} = n$

Square both sides (noting that this may introduce spurious solutions):

$12 - n = {n}^{2}$

This derived equation may (and actually does) have solutions which are not solutions of the original equation.

Subtract $\left(12 - n\right)$ from both sides to get:

$0 = {n}^{2} + n - 12 = \left(n + 4\right) \left(n - 3\right)$

Hence $n = - 4$ or $n = 3$

$n = - 4$ is not a solution of the original equation since:

$\sqrt{12 - \left(- 4\right)} = \sqrt{16} = 4 \ne - 4$

$n = 3$ is a valid solution of the original equation:

$\sqrt{12 - 3} = \sqrt{9} = 3$