# How to find the number of solutions for $|x^2 - 4| = c$, is this just guess work, if so whats the approch?

I'm stuck on a problem from the precalculus book it's an extra problem not part of the main exercise, but here is the question

Q - if $$|x^2 - 4| = c$$ find the value of $$c$$ For which there are $$4,3,2,0$$, solutions ( this is $$\#$$ of solutions) and can there be more, number of solutions $$(1)$$ if yes, write them down.

How do I approach this?

• May be the graph of $y=\vert x^2-4\vert$ would help you the see it geometrically. Commented Nov 3, 2023 at 3:59
• @Bumblebee one can do that way, but I was looking for a more analytic approach rather than a graphical, if possible Commented Nov 3, 2023 at 4:03

Our cases for this solution depend on the value of $$c$$. The first case would be when $$c<0$$, one case when $$c=0$$, one case when $$0, one case when $$c=4$$ and so on.

Firstly we note (when $$c<0$$) there are no solutions as $$|x^2-4| \ge 0$$ for all $$x$$. Let's proceed to the next case. So (if $$f(x)=x^2-4$$ and $$g(x)=|f(x)|$$) then $$f$$ has a minimum at $$(0, -4)$$. Hence $$g$$ has a local maximum at $$(0,4)$$. Can you draw the graph of $$g(x)$$?

It thus follows, if $$0=c$$ there are $$2$$ solutions to $$|x^2-4|=c$$, if $$0, there would be $$4$$ solutions to $$|x^2-4|=c$$, if $$c=4$$ there would be (why?) $$3$$ solutions to $$|x^2-4|=c$$ and lastly, if $$c>4$$, can you write down (using a graph) the number of solutions to $$|x^2-4|=c$$?

Thanks.

• I guess knowing the behaviour of absolute value functions and quadratic functions is what it takes to solve this question Commented Nov 3, 2023 at 4:33
• True, if you have a graphing software (like Wolfram Alpha) - then plot this function on it, and maybe a few different ones(?) Commented Nov 3, 2023 at 11:14

Consider first expressing all the possible solutions in a general form, in terms of $$c$$. Then analyse the domain w.r.t $$c$$ for each of the different solutions, such that $$x$$ is real.

Your equation can have a maximum of $$4$$ solutions:

$$|x^2 - 4| = c \iff \sqrt{(x^2 - 4)^2} = c \iff (x^2 - 4)^2 = c^2 \\\therefore x^2 - 4 = \pm c \iff x = \pm\sqrt{4\pm c}$$

So we have

$$x = \begin{cases} \sqrt{4 + c} & c \ge -4 \\-\sqrt{4 + c} & c \ge -4 \\\sqrt{4 - c} & c \le 4 \\-\sqrt{4 - c} & c \le 4 \end{cases}$$

Now, observe for the particular subset of reals, where $$c \in (-4, 4)$$ all $$4$$ solutions are real (exist), while unique values such as $$c = \pm 4$$ takes on double roots; $$\sqrt{4 - (4)} = 0 = -\sqrt{4 - (4)}$$, hence $$3$$ possible solutions. For any other $$c$$, only $$2$$ solutions are real.