## Task A-2.1.

Determine all real numbers $x$ for which

$$
\sqrt{|x-1|+|x+4|}=|x|
$$
## Solution.

By squaring the initial equality, we obtain that

$$
|x-1|+|x+4|=x^{2}
$$

We notice that the left side of the equation takes different values when $x \leqslant-4$, when $-4 < x \leq 1$, and when $x > 1$, and we consider each case separately.

In the case $x \leqslant-4$, for the expression on the left side of the equality, we have

$$
|x-1|+|x-4|=(1-x)+(-4-x)=-3-2 x
$$

This gives us the equation

$$
-3-2 x=x^{2}
$$

or

$$
x^{2}+2 x+3=0
$$

The discriminant of the obtained quadratic equation is $D=-8$. For the expression on the left side of the equality, we have

$$
|x-1|+|x-4|=(x-1)+(x-4)=x^{2}
$$

thus we get

$$
2 x+3=x^{2}
$$

or

$$
0=x^{2}-2 x-3
$$

The solutions to this quadratic equation are $x=3$ or $x=-1$. From the condition $x>1$, we conclude that the only possible solution is $x=3$.

By directly checking $x=3$ in the initial equality, we see that it is indeed a solution.

Finally, all real numbers $x$ that satisfy the given equation are $x=-\sqrt{5}$ and $x=3$.

Note: The equality in the problem statement and the first equation in the solution obtained by squaring it are actually equivalent equations. In the initial equality, both sides are non-negative, so any solution to the equation

$$
|x-1|+|x+4|=x^{2}
$$

is also a solution to the equation

$$
\sqrt{|x-1|+|x+4|}=\sqrt{x^{2}}=|x| .
$$

The reverse statement obviously holds. In solutions that express this statement, it is not necessary to subsequently check that the numbers $x=-\sqrt{5}$ and $x=3$ are indeed solutions to the initial equation, and the sixth and tenth points from the above scoring scheme are immediately awarded upon finding the corresponding values for $x$.