For which values of $a$ does $||x^2-16|-7|-2a=0$ have 3 roots? Here is the equation:
$$||x^2-16|-7|-2a=0$$
Please help solve this.
For which $a$ does the equation have 3 roots?
 A: Hint: If $x$ is solution then so is $-x$. So one of the solutions should definitely be $0$, otherwise you would get even number of roots
A: Api, first, use the definition of absolute value.
$||x^2-16|-7|-2a=0$ implies that two cases:
1) $|x^2-16|-7=2a$
This equation gives us $a=\frac{x^2-23}{2} $ and $a=\frac{x^2-9}{2}$.
2) $|x^2-16|-7=-2a$
Again, this equation gives us $a=\frac{23-x^2}{2} $ and $a=\frac{x^2+9}{2}$.
We should have the following this case.
$x=0$ and $x=\pm x_0$.
Hence, if $x=0$, then the only possible case is $a=\frac{9}{2}$.
Why?
if put $0$ in the first equation in the second case, then you get $a=\frac{23}{2}$ and it implies that $x\geq 4$ and a contradiction. The rest is similar.  
A: There is also a simplier, graphical solution to it.
Every time when you plot a graph involving the absolute thing, you can just "flip" the negative region upwards. As you can see in the graph, only when $2a=9$, that is $a=9/2$ will exactly 3 solutions exist. 

This method is typically useful when you are asked to find the no. solution for $y=0$.
Hope it helps.
