Question

I need help with my algebra 2. I also want to know how to solve this.

Answer 1

Add 1 to both sides:

$\sqrt{x+3} = x+1$

In cases like this, we have to remember that a root is always positive, so we can square both sides only assuming that

$x+1 \geq 0 \iff x \geq -1$

Under this assumption, we square both sides and we have

$x+3 = (x+1)^2 \iff x+3 = x^2+2x+1 \iff x^2+x-2 = 0$

The solutions to this equation are

$x = -2,\ x=1$

But since we can only accept solutions greater than -1, we discard $x=-2$ and accept $x=1$.

In fact, we have

$x=-2 \implies \sqrt{-2+3}-1=0\neq -2$

and

$x=1 \implies \sqrt{4}-1=1$

which is the only solution.