Solve for $x\in\mathbb{R}$: $x^{2} + 2|x-3| - 10 \leq 0$

Question

I went about taking cases for $x^{2} + 2|x-3|- 10\leq 0$.

Taking $x-3$ and $-(x-3)$ as cases. Is it the correct approach?

Taking cases I realise I get $x^{2} + 2x - 16 \leq 0$ and $x^{2}-2x-4 \leq 0$.

How would I take intersection of solutions here?

Answer 1

I like the following way.

We need to solve $$2|x-3|\leq10-x^2,$$ which is $$-(10-x^2)\leq2(x-3)\leq10-x^2,$$ which is $$x^2+2x-16\leq0$$ and $$x^2-2x-4\leq0,$$ which is $$-1-\sqrt{17}\leq x\leq-1+\sqrt{17}$$ and $$1-\sqrt{5}\leq x\leq1+\sqrt5,$$ which gives the answer: $$[1-\sqrt5,\sqrt{17}-1].$$

Answer 2

For $$x\geq 3$$ we have to solve $$x^2+2(x-3)-10\le 0$$ this is fulfiiled for $$3\le x\le -1+\sqrt{17}$$ For $$x<3$$ we have to solve $$x^2-2(x-3)-10\le 0$$ this gives $$1-\sqrt{5}\le x<3$$