Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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


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.

share|cite|improve this question
You should draw the graph of the function $f(x)=\vert \vert x\vert -1\vert$. – Etienne Mar 10 '14 at 19:36
up vote 2 down vote accepted

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 $a\in(0,1)$.

share|cite|improve this answer
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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.