For which values of the constant $c$ does the equation

$$
x^{2}-2\left|x+\frac{1}{4}\right|+c=0
$$

have exactly three distinct real roots?
Solution. "Break down" the absolute value sign:

I. case: $x \geq-\frac{1}{4}$; the equation then becomes $x^{2}-2 x+c-\frac{1}{2}=0$, the possible values of the solutions are:

$$
x_{1}=1+\sqrt{\frac{3}{2}-c} \quad \text { and } \quad x_{2}=1-\sqrt{\frac{3}{2}-c}
$$

$I / A$ : There are two different real roots if $c<\frac{3}{2}$ and $c \geq-\frac{1}{16}
$$

$I / B$ : There is exactly one real root precisely when $c=\frac{3}{2}$ (when $x_{1}=1 \geq-\frac{1}{4}$ ) or

$$
c<-\frac{1}{16}
$$

$I / C$ : There are no real roots precisely when $c>\frac{3}{2}$.

II. case: $x<-\frac{1}{4}$; the equation then becomes $x^{2}+2 x+c+\frac{1}{2}=0$, the possible values of the solutions are:

$$
x_{1}=-1+\sqrt{\frac{1}{2}-c} \quad \text { and } \quad x_{2}=-1-\sqrt{\frac{1}{2}-c}
$$

$II / A$ : There are two different real roots if $c<\frac{1}{2}$ and $c \geq-\frac{1}{16}
$$

$II / B$ : There is exactly one real root precisely when $c=\frac{1}{2}$ (when $x_{1}=-1 \geq-\frac{1}{4}$ ) or

$$
c<-\frac{1}{16}
$$

$II / C$ : There are no real roots precisely when $c>\frac{1}{2}$.

For the equation to have exactly 3 different real roots, the following cases can lead to this:

(i) I/A and II/A hold, and one of the two roots from each case coincides;

(ii) I/A and II/B hold, and the solution of the latter is different from both roots of the former;

(iii) I/B and II/A hold, and there is no coincidence among the roots of the two types.

In case (i), $\frac{3}{2}>c \geq-\frac{1}{16}$ and $\frac{1}{2}>c>-\frac{1}{16}$, so $-\frac{1}{16}<c<\frac{1}{2}$. The only possible cases are:

$$
1-\sqrt{\frac{3}{2}-c}=-1+\sqrt{\frac{1}{2}-c} \quad \text { or } \quad 1-\sqrt{\frac{3}{2}-c}=-1-\sqrt{\frac{1}{2}-c}
$$

Rearranging and squaring both sides in both cases yields the same result:

$$
\begin{gathered}
2-\sqrt{\frac{3}{2}-c}=+\sqrt{\frac{1}{2}-c}, \quad \text { or } \quad 2-\sqrt{\frac{3}{2}-c}=-\sqrt{\frac{1}{2}-c} \\
4+\frac{3}{2}-c-\sqrt{24-16 c}=\frac{1}{2}-c \\
5=\sqrt{24-16 c}, \quad 25=24-16 c, \quad c=-\frac{1}{16}
\end{gathered}
$$

which is impossible in this case.

In case (ii), $\frac{3}{2}>c \geq-\frac{1}{16}$, and $c=\frac{1}{2}$ or $c \leq-\frac{1}{16}$, so $c=\frac{1}{2}$ or $c=-\frac{1}{16}$. If $c=\frac{1}{2}$, the roots are: 2, 0, and $-1$; if $c=-\frac{1}{16}$, the roots are $\frac{9}{4}, -\frac{1}{4}$, and $-\frac{7}{4}$; both cases are valid.

In case (iii), $c=\frac{3}{2}$ or $c<-\frac{1}{16}$, and $-\frac{1}{16}<c<\frac{1}{2}$, this case cannot occur.

Therefore, there are exactly 3 different real roots if $c=\frac{1}{2}$ or $-\frac{1}{16}$.