Inequality with roots, a step is not clear!

Question

$$\sqrt{x+3}\le{x+2}$$

$\ x+3\le{x^2+4x+4}$ --> $\ -x^2-4x+4+x+3\le0$ --> $\ x^2+4x+4-x+3\ge0$ --> $\ x^2+3x+1\ge0$

-->

$$\ (-b\pm\sqrt{b^2-4ac})/2a$$

-->

$$\ (-3\pm\sqrt{5})/2a$$

-->

The solution is:

$$\ (-\infty, (-3\pm\sqrt{5})/2] U [(-3\pm\sqrt{5})/2, +\infty$$

FROM HERE

Divide the inequality in the 2 possible cases:

a. $\ \sqrt{x+3}\le{x+2}$, $\ x+2\ge0$

b. $\ \sqrt{x+3}\le{x+2}$, $\ x+2\lt0$

a)--> same resolution as before, and, $\ x\ge-2$

b)--> since the first component os always $\ge0$: $\ x\in\varnothing$, $\ x\lt-2$

THEN THE CONCLUSION IS OBVIOUS.

I'd like to know what is the general rule to apply with "Divide with 2 possible cases".

1. Why do I have to?
2. How do I know when to do this?

I made everything correct but basically didn’t do half of the exercise :/ from HERE is where starts the part i left.

Thanks to everyone will respond :)

Answer 1

When working with inequalities, I always look at the domain first, it helps me to eliminate unnecessary scenarios. Right off the bat we can restrict the interval of possible solutions to $x+3 \ge 0$ (expression under the root should be non-negative) and $x+2 \ge 0$ (root should be non-negative) or $x \ge -2$. Keeping this in mind, we can square both parts and solve the inequality without fear that we will obtain false solutions.

Answer 2

We have $$x\geq -3$$ and $$x\geq -2$$ so $$x\geq -2$$ After squaring we get $$0\le x^2+3x+1$$ and this is $$0\le (x+\frac{3}{2}+\frac{\sqrt{5}}{2})(x+\frac{3}{2}-\frac{\sqrt{5}}{2})$$ and from here we get $$\frac{\sqrt{5}}{2}-\frac{3}{2}\le x<\infty)$$