$ x-b = 4 \left | 4 | x | -b ^ 2 \right | $ has exactly three solutions.

Question

Find the sum of all values of the parameter $ b $ for which the equation $$ x-b = 4 \left | 4 | x | -b ^ 2 \right | $$ has exactly three solutions.

I tried to represent this equation in two axes and draw this graph in order to analyze the cases when there are exactly three solutions, but it comes out too cumbersome. I will be glad for any hint, thanks :)

Answer 1

Just reminding some basics:

• To plot $f(|x|)$, draw $f(x)$, ignore the left hand side part of the y-axis, reflect the graph to the right of the y-axis in the y-axis.
• To plot $|f(x)|$, draw $f(x)$, reflect any part of graph that was below the x-axis in the x-axis.

Let's draw $4|4|x| - b^2| = |16|x| - 4b^2|$:

1. $f(x) = 16x - 4b^2$.
2. $f(|x|) = 16|x| - 4b^2$
3. $|f(|x|)|$ = $|16|x| - 4b^2|$

Then we have a $y = x - b$, a line parallel to lines below (or one of the lines below):

In order for $x - b = 4|4|x| - b^2|$ to have exactly three answers, $x - b$ should have three intersections with $4|4|x| - b^2|$. So $x - b$ should be one of the blue lines below and can't be any line else:

So:

$$ 4b^2=-b \Rightarrow \begin{cases} b=0\\ b=-4\\ \end{cases} $$

$$ \frac{-b^2}{4}=b \Rightarrow \begin{cases} b=0\\ b=\frac{-1}{4}\\ \end{cases} $$

But if $b=0$, then $\frac{b^2}{4}=\frac{-b^2}{4}$, so the graph we drew will change and the only answer will be $x=0$.

So $ \begin{cases} b=-4 \begin{cases} x=-4\\ x=\frac{60}{17}\\ x=\frac{68}{15}\\ \end{cases} \\ b=\frac{-1}{4} \begin{cases} x=\frac{-1}{34}\\ x=0\\ x=\frac{1}{30}\\ \end{cases} \\ \end{cases} $