Problem 8.1. Solve the equation

$$
|| x-\frac{5}{2}\left|-\frac{3}{2}\right|=\left|x^{2}-5 x+4\right|
$$

Ivan Tonov
8.1. Since both sides of the equation are non-negative, it is equivalent to

$$
\left(\left|x-\frac{5}{2}\right|-\frac{3}{2}\right)^{2}=\left(x^{2}-5 x+4\right)^{2}
$$

Hence

$$
\left(\left|x-\frac{5}{2}\right|-\frac{3}{2}-x^{2}+5 x-4\right)\left(\left|x-\frac{5}{2}\right|-\frac{3}{2}+x^{2}-5 x+4\right)=0
$$

and we have to solve the equations

$$
\left|x-\frac{5}{2}\right|-x^{2}+5 x-\frac{11}{2}=0
$$

and

$$
\left|x-\frac{5}{2}\right|+x^{2}-5 x+\frac{5}{2}=0
$$

For $x \leq \frac{5}{2}$ we have respectively that $\frac{5}{2}-x-x^{2}+5 x-\frac{11}{2}=0 \Longleftrightarrow x^{2}-$ $4 x+3=0$, i.e. $x=1$ or $x=3$ and $\frac{5}{2}-x+x^{2}-5 x+\frac{5}{2}=0 \Longleftrightarrow x^{2}-6 x+5=0$, i.e. $x=1$ or $x=5$. Therefore $x=1$ is the only solution of the given equation in this case.

Analogously, for $x>\frac{5}{2}$ we have $x-\frac{5}{2}-x^{2}+5 x-\frac{11}{2}=0 \Longleftrightarrow x^{2}-6 x+8=$ 0 , i.e, $x=2$ or $x=4$ and $x-\frac{5}{2}+x^{2}-5 x+\frac{5}{2}=0 \Longleftrightarrow x^{2}-4 x=0$, i.e. $x=0$ or $x=4$. Therefore $x=4$ is the only solution of the given equation in this case.

Hence the given equation has two solutions $x_{1}=1$ and $x_{2}=4$.