# Real parameter equation

I'm having a problem with this question:
For which values of the real parameter a the equation:

$$||x|-1|=a$$

has exactly 4 solutions?
The solution is this: $$0 < a < 1$$

What I tried was this:

1) Make the first condition, $a>0$,
Split the equation in these two: $(1)~|x|-1=a;~(2)~|x|-1=-a$.
2) Make the second conditions, $(1)~a+1>0,~a>-1; (2)~-a+1>0,~a<1$
If the conditions are true, the equation will have 4 solutions, therefore, a is contained in the interval $<0,1>$, which is $0 < a < 1$.

Was the described procedure correct?

Thanks in advance.

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You should draw the graph of the function $f(x)=\vert \vert x\vert -1\vert$. – Etienne Mar 10 '14 at 19:36

## 1 Answer

Notice that $$|x|=\alpha\iff x=\pm\alpha$$ so we have two values of $x$ if $\alpha>0$ hence

$$||x|-1|=a\iff |x|-1=\pm a\iff|x|=1+\pm a\iff x=\pm(1+\pm a)$$ and we have $4$ values of $x$ if $a>0$ and $1+\pm a>0$ hence $a>-1$ and $a<1$ so $a\in(0,1)$.

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So I did it the right way... – user134480 Mar 10 '14 at 19:47
Yes you did it. – user63181 Mar 10 '14 at 19:49