For which values of the parameters $a$ and $b$ is the solution of the following equation real? Provide some pairs of $a, b$ values to numerically illustrate the different cases.

$$
\sqrt{2 a+b+2 x}+\sqrt{10 a+9 b-6 x}=2 \sqrt{2 a+b-2 x}
$$
We move the first term on the left side to the right side next to the similar term, square the equation, and rearrange:

$$
\sqrt{(2 a+b)^{2}-4 x^{2}}=-b
$$

From this, we immediately obtain a necessary condition for the existence of a solution:

$$
b \leqq 0
$$

By squaring again, we get $x^{2}=a^{2}+a b$, and thus the only possible solutions are:

$$
x_{1}=\sqrt{a(a+b)}, \quad x_{2}=-\sqrt{a(a+b)}
$$

These are real only if $a$ and $a+b$ have the same sign, or if one of them is zero. Combining this with (2), $x_{1,2}$ are real if

$$
a \geqq a+b \geqq 0, \text { i.e., } -a \leqq b \leqq 0
$$

or if

$$
a+b \leqq a \leqq 0, \text { i.e., } a \leqq 0, b \leqq 0
$$

Of course, we can only accept a solution that makes each of the three square roots in (1) real. We see that this is not satisfied in case (5), because substituting either $x_{1}$ or $x_{2}$, at least one of the first and third square roots will have a negative number under it, except when $2 a+b=0$ and $a(a+b)=0$, which necessarily leads to $a=b=0$. This is already included in condition (4), so we can exclude case (5).

In case (4), the numbers under the first and third square roots are non-negative, since dividing them by 2, we get the arithmetic and geometric means of the non-negative numbers $a$ and $a+b$. Similarly, the number under the second square root is also non-negative, since

$$
\frac{10 a+9 b}{6}=\frac{a+9(a+b)}{6}=\frac{1}{2}\left\{\frac{a}{3}+3(a+b)\right\} \leqq \sqrt{\frac{a}{3} \cdot 3(a+b)}=\left|x_{1,2}\right|
$$

We only need to check which of (3) - and under what additional conditions beyond (4) - satisfies (1). Taking $x_{1}$, the first and third square roots become:

$$
\begin{gathered}
\sqrt{2 a+b+2 x_{1}}=\sqrt{a+2 \sqrt{a(a+b)}+(a+b)}=\sqrt{a}+\sqrt{a+b} \\
\sqrt{2 a+b-2 x_{1}}=\sqrt{a}-\sqrt{a+b}
\end{gathered}
$$

Thus, subtracting the first term on the left side from the right side, we get

$$
\sqrt{a}-3 \sqrt{a+b}
$$

The second square root, however, is

$$
\sqrt{10 a+9 b-6 x_{1}}=\sqrt{a-6 \sqrt{a(a+b)}+9(a+b)}= \pm(\sqrt{a}-3 \sqrt{a+b})
$$

depending on whether

$$
\begin{array}{ll}
a \geqq 9(a+b), \text { i.e., } b \leqq-\frac{8}{9} a \leqq 0, & \text { or } \\
a<9(a+b), \text { i.e., }-\frac{8}{9} a<b \leqq 0
\end{array}
$$

Accordingly, (1) is satisfied under (4a), but not under (4b).

For $x_{2}$, the roles of the first and third square roots are reversed, and only the second square root needs to be recalculated, which is similarly

$$
\sqrt{a}+3 \sqrt{a+b}
$$

and with it, (1) is satisfied.

Therefore, the solution is

$$
\begin{array}{ll}
-a<b \leqq-\frac{8}{9} a<0 \text { case } & x= \pm \sqrt{a(a+b)} \\
-\frac{8}{9} a<b \leqq 0 \text { case } & x=-\sqrt{a(a+b)}
\end{array}
$$

In the case $a \geqq 0, a+b=0$, $x=0$, so the number of solutions is 2, 1, or 0 depending on the two conditions.

Examples:

For $a=16, b=-7$, $x_{1}=12$ is not a solution: $7+5 \neq 2$,

$$
x_{2}=-12 \text { is a solution }: \quad 1+13=14
$$

For $a=16, b=-15(<-128 / 9)$,

$x_{1}=4$ and $x_{2}=-4$ are both solutions: $5+1=6, \quad 3+7=10$;

For $a=1, b=3$, neither $x_{1}=2$ nor $x_{2}=-2$ is a solution: $3+5 \neq 2,1+7 \neq 6$.

Szendrei Mária (Szeged, SÁgvári E. Gyak. Gimn., III. o. t. )