Solve the following equation:

$$
\sqrt{2 x+3}+\sqrt{5 x+1}=\sqrt{12 x+13} .
$$
In the equation, by convention, the square roots always represent only one value, the positive one. Squaring both sides:

$$
2 x+3+5 x+1+2 \sqrt{(2 x+3)(5 x+1)}=12 x+13
$$

After simplification:

$$
2 \sqrt{(2 x+3)(5 x+1)}=5 x+9 \ldots
$$

Emphasizing that $5 x+9>0$, we square again:

$$
40 x^{2}+68 x+12=25 x^{2}+90 x+81 \quad \text { or } \quad 15 x^{2}-22 x-69=0 \ldots
$$

The discriminant of this equation is: $4(121+69 \cdot 15)=4 \cdot 1156=(2 \cdot 34)^{2}$.

The roots of (2) are:

$$
x_{1}=\frac{22+68}{30}=3, \quad x_{2}=\frac{22-68}{30}=-\frac{23}{15}
$$

We need to check whether these values satisfy the original irrational equation.

$x_{1}=3$ meets the condition $5 x+9>0$ and also the requirement that the expressions under the square root in the original equation are positive:

$$
\sqrt{2 \cdot 3+3}=\sqrt{9}=+3, \quad \sqrt{5 \cdot 3+1}=\sqrt{16}=+4, \quad \sqrt{12 \cdot 3+13}=\sqrt{49}=+7
$$

Indeed,

$$
3+4=7
$$

$x_{2}=-\frac{23}{15}$ meets the condition $5 x+9>0$, but the expressions under the square root in the original equation are negative:

$$
2 x_{2}+3=-\frac{1}{15}, \quad 5 x_{2}+1=-\frac{100}{15}, \quad 12 x_{2}+13=-\frac{81}{15}
$$

Thus, the square roots are imaginary, and since

$$
-i \sqrt{\frac{1}{15}}+10 i \sqrt{\frac{1}{15}}=9 i \sqrt{\frac{1}{15}} \cdots
$$

$x_{2}$ satisfies the equation

$$
-\sqrt{2 x+3}+\sqrt{5 x+1}=\sqrt{12 x+13} \ldots
$$

($x_{2}$ is an extraneous root!)

Jakab Károly (Catholic High School 6th grade private student, Kalocsa).

Note. We have pointed out several times that squaring an irrational equation with square roots leads to a rational equation whose roots are not necessarily roots of the original equation. Therefore, in each case, we must verify by substitution whether the obtained root values satisfy the original equation. We can also determine which equation the extraneous root is a root of.

If we think about solving the equation graphically, then

$$
\sqrt{2 x+3}, \quad \sqrt{5 x+1}, \quad \sqrt{12 x+13}
$$

can only represent real values, so it must be

$$
2 x+3 \geq 0, \quad 5 x+1 \geq 0, \quad 12 x+13 \geq 0
$$

i.e., the conditions $x \geq-\frac{3}{2}, x \geq-\frac{1}{5}, x \geq-\frac{13}{12}$ must be satisfied. This happens when $x \geq-\frac{1}{5}$.

From this perspective, the equation (4) mentioned in the solution has no significance.

In some solutions, the statement appears that "since the original equation is linear, it can only have one solution." If the equation involves a rational function of $x$ or an irrational function of $x$, we do not speak of a degree.

In some solutions, there is an attempt to prove that $x_{2}=-\frac{23}{15}$ also satisfies the equation. Substituting this value into the original equation, we get:

$$
\sqrt{-\frac{1}{15}}+\sqrt{-\frac{20}{3}}=\sqrt{-\frac{27}{5}}
$$

This is correct so far. Squaring, we indeed get: $-\frac{1}{15}+2 \sqrt{\frac{20}{45}}-\frac{20}{3}=-\frac{27}{5}$.

There are two errors here. First: if $a=b$, then $a^{2}=b^{2}$. But if $a^{2}=b^{2}$, then $a= \pm b$ follows.

Secondly:

$$
\sqrt{-\frac{1}{15}}=i \sqrt{\frac{1}{15}}, \quad \sqrt{-\frac{20}{3}}=i \sqrt{-\frac{20}{3}}
$$

so their product is:

$$
i^{2} \sqrt{-\frac{20}{45}}=-\sqrt{\frac{20}{45}}
$$

In other words: the rule for multiplying square roots, which states that "we multiply under the square root," is not valid when the square roots represent imaginary numbers.