# How do you solve sqrt(2n+3)=n and check your solution?

Mar 25, 2018

$n = 3 , n = - 1$

#### Explanation:

Square both sides:

${\sqrt{2 n + 3}}^{2} = {n}^{2}$

Recall that ${\sqrt{a}}^{2} = a$. Thus,

$2 n + 3 = {n}^{2}$

Get this into standard quadratic form, meaning we move everything to one side. We want ${n}^{2}$ to stay positive, so we'll move everything to the right.

${n}^{2} - 2 n - 3 = 0$

Solve this for $n .$ Easiest way is by factoring. Two numbers that add to give $- 2$ and multiply to give $- 3$ are $- 3 , 1.$

$\left(n - 3\right) \left(n + 1\right) = 0$

So,

$n = 3 , n = - 1$

Let's plug these into the square root to ensure we don't get a negative root, as negative roots will not give us a solution we desire.

$\sqrt{2 \left(3\right) + 3} > 0$, $n = 3$ is fine.

$\sqrt{2 \left(- 1\right) + 3} > 0 ,$ $n = - 1$ is fine.