## Task A-2.1.

Determine all real solutions of the equation

$$
\sqrt{x^{2}+x+3}-\sqrt{x^{2}-2 x+4}=1
$$
## Solution.

By moving one expression from the left side of the equation to the right and squaring, we get

$$
\begin{array}{cc}
-\sqrt{x^{2}-2 x+4}=1-\sqrt{x^{2}+x+3} & \\
x^{2}-2 x+4=1-2 \sqrt{x^{2}+x+3}+x^{2}+x+3 & 1 \text { point } \\
-3 x=-2 \sqrt{x^{2}+x+3} & 3 \text { points } \\
3 x=2 \sqrt{x^{2}+x+3} &
\end{array}
$$

By squaring again and rearranging, we have

$$
\begin{gathered}
9 x^{2}=4\left(x^{2}+x+3\right) \\
9 x^{2}=4 x^{2}+4 x+12 \\
5 x^{2}-4 x-12=0
\end{gathered}
$$

The solutions to this quadratic equation are \( x_{1}=2 \) and \( x_{2}=-\frac{6}{5} \).

By substituting the solution \( x=2 \) into the original equation, we see that it is a solution. 1 point

On the other hand, by substituting \( x=-\frac{6}{5} \) into the original equation, we see that it is not a solution. We conclude that the only solution to the original equation is \( x=2 \).

Note: That \( x=-\frac{6}{5} \) is not a solution to the original equation can also be seen by the fact that it definitely does not satisfy the equality \( 3 x=\sqrt{x^{2}+x+3} \), as one side of the equality is positive and the other is negative.

The solution can be carried out (and scored) analogously even if the term \( \sqrt{x^{2}+x+3} \) is not moved to the right side of the equation, but \( \sqrt{x^{2}-2 x+4} \) is.