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

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Answer 1 · 2018-03-25T20:46:56

$n=3, n=-1$

Square both sides:

$sqrt(2n+3)^2=n^2$

Recall that $sqrt(a)^2=a$ . Thus,

$2n+3=n^2$

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.

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

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

$(n-3)(n+1)=0$

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.

$sqrt(2(3)+3)>0$ , $n=3$ is fine.

$sqrt(2(-1)+3)>0,$ $n=-1$ is fine.

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