# How do you solve y=sqrt(12-y) and check your solution?

Apr 20, 2017

See below.

#### Explanation:

Let's start by squaring both sides.

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

${y}^{2} + y - 12 = 0$

Factoring,

$\left(y + 4\right) \left(y - 3\right) = 0$

So $y = - 4 , 3$

We can check to see which of these solutions work.

If $y = - 4$, then:

$- 4 \setminus \ne \sqrt{12 - \left(- 4\right)}$, as a square root does not come out to be negative (in real numbers).

Thus, $y = - 4$ is extraneous, and is not a solution.

Now let's try $y = 3$

$3 = \sqrt{12 - 3} = \sqrt{9} = 3$ is true.

The only solution is therefore $y = 3$.