For which values of $a$ the equation $4x-\left|3x-|x+a|\right|=9|x-1|$ has two answers?

Question

For which values of $a$ the equation $4x-\left|3x-|x+a|\right|=9|x-1|$ has two roots?

I wrote the equation as $$4x-9|x-1|=\left|3x-|x+a|\right|$$ We have $4x-9|x-1|\ge0$. Hence $\frac9{13}\le x\le \frac95$. So I found domain of $x$.

To continue, I tried squaring both side of the equation but it makes things more complicated.

Answer 1

Let $ f(x) = 4x - | 3x - |x+a| |$ and $ g(x) = 9 |x-1 | $.

Hints: (If you're stuck, show what you've tried.)

1. Show that $f(x)$ is non-decreasing.
2. Show that slope of $f(x)$ (except at kink points) is $ 4 \pm 3 \pm 1 = \{ 0, 2, 4, 8 \}$.
3. Show that slope of $ g(x) $ is either -9 or 9. What are the corresponding domains?
4. Show that on $ x \in ( -\infty , 1 ]$, there is at most 1 solution to $ f(x) = g(x)$. What is a (simple) necessary and sufficient condition (NASC) for there to be 1 solution?
5. Show that on $ x \in (1, \infty)$, there is at most 1 solution to $ f(x) = g(x)$. Show that the NASC from before is still a necessary and sufficient condition for there to be 1 solution.
6. Hence, if there are 2 solutions to $f(x) = g(x)$, then NASC holds.
7. Hence, conclude what the values of $a$ are.

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The NASC that I'm thinking of is

$f(1) > g(1) = 0 $.
You still have to show that this is necessary and sufficient.