How to find the intersection of the graphs of $y= x^2$ and $y = 6 - |x|$?

Question

I was trying to solve this question in preparation for the Math subject GRE exam:

The region bounded by the graphs of $y=x^{2}$ and $y=6-|x|$ is revolved around the $y$ -axis. What is the volume of the generated solid?
(A) $\frac{32}{3} \pi$
(B) $9 \pi$
(C) $8 \pi$
(D) $\frac{20}{3} \pi$
(E) $\frac{16}{3} \pi$

So first I was trying to find the intersection of the graphs of $y= x^2$ and $y = 6 - |x|.$ I did that by equating $ x^2 = 6 - |x|.$ Then I ended up having 2 equations which are $ x^2 - 6 + x = 0 $ and $ x^2 - 6 - x = 0 $, and then I got four values of $x$, which are $\pm 3, \pm 2.$ But I found in the answer for the question that they consider only the two values $\pm 2.$

Could anyone show me what is wrong in my solution or in the book solution (the book just gave me the values $\pm 2$ without explaining how and why), please?

Answer 1

The solutions should be ($x^2-6+x=0$ and $x\ge0$) or ($x^2-6-x=0$ and $x\le0)$.

Can you take it from here?

Answer 2

No need to define cases: as $|x|^2=x^2$, you only need to find the nonnegative roots of the equation $$ t^2+t-6=0.$$ (Vieta's relations show it has a positive and a negative root) and deduve the values of $x$.