Solve the following equation:

$$
3 x^{2}-a \sqrt{a^{2}-x^{2}}-2 a^{2}=0
$$
I. Solution. The square root in the equation is real if $x^{2} \leq a^{2}$. Therefore, we can only consider the solution of the equation that satisfies this requirement.

The equation can be written in the form

$$
3 x^{2}-2 a^{2}=a \sqrt{a^{2}-x^{2}} \ldots
$$

and assume that $a>0$. Then it must be:

$$
3 x^{2}-2 a^{2}>0 \quad \text { that is } \quad a^{2} \geq x^{2}>\frac{2 a^{2}}{3} \cdots
$$

Squaring both sides of equation 1) and rearranging,

$$
9 x^{4}-11 a^{2} x^{2}+3 a^{4}=0
$$

and from this $\alpha=\left(x^{2}\right)_{1}=\frac{11+\sqrt{13}}{18} a^{2}, \beta=\left(x^{2}\right)_{2}=\frac{11-\sqrt{13}}{18} a^{2}$.

Of these, only $\alpha=\frac{11+\sqrt{13}}{18} a^{2}$ satisfies condition 2) and thus the given equation is satisfied by

$$
x= \pm \frac{a}{3} \sqrt{\frac{11+\sqrt{13}}{2}} \text { . }
$$

The value denoted by $\beta$ is a root of the equation $3 x^{2}+a \sqrt{a^{2}-x^{2}}-2 a^{2}=0$.

If $a<0$, the solution should be considered with the opposite sign. In this case,

$$
x^{2}=a^{2}-y^{2}
$$

Substituting into the given equation:

$$
3\left(a^{2}-y^{2}\right)-a y-2 a^{2}=0, \quad \text { i.e. } \quad 3 y^{2}+a y-a^{2}=0
$$

and

From this, since $y>0, y=\frac{-a+\sqrt{a^{2}+12 a^{2}}}{6}=\frac{\sqrt{13}-1}{6} a$, provided $a>0$

$$
x^{2}=a^{2}-\left(\frac{\sqrt{13}-1}{6}\right)^{2} a^{2}=\frac{11+\sqrt{13}}{18} a^{2}
$$

thus

$$
x= \pm \frac{a}{3} \sqrt{\frac{11+\sqrt{13}}{2}} \sim \pm 0.9 a
$$

Iván Steiner (Ferenc Toldy r. V. o. Bp. II.)

Supplement. If $a<0$, the case where $y<0$ will arise, when

$$
y=\frac{-\sqrt{13}-1}{6} a=-\frac{\sqrt{13}+1}{6} a
$$

and

$$
\begin{aligned}
& x^{2}=a^{2}-\left(\frac{\sqrt{13}+1}{6}\right)^{2} a^{2}=\frac{11-\sqrt{13}}{18} a^{2} \\
& x= \pm \frac{a}{3} \sqrt{\frac{11-\sqrt{13}}{2}} \sim \pm 0.64 a
\end{aligned}
$$