## Task 2 - 301212

Determine all real numbers $a$ for which the equation

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

has two real solutions, which, when denoted in a suitable order as $x_{1}$ and $x_{2}$, satisfy the condition

$$
6 x_{1}+x_{2}=0
$$
For every real number $a$, $\frac{a^{2}}{36}+\frac{2}{3}>0$. Therefore, (1) has for every real number $a$ two real solutions, namely the two numbers

$$
-\frac{a}{6} \pm \sqrt{\frac{a^{2}}{36}+\frac{2}{3}}=\frac{1}{6}\left(-a \pm \sqrt{a^{2}+24}\right)
$$

I. If these two numbers satisfy condition (2), then either

$$
\begin{gathered}
\left(-a+\sqrt{a^{2}+24}\right)+\frac{1}{6}\left(-a-\sqrt{a^{2}+24}\right)=0 \quad \text { or } \\
\left(-a-\sqrt{a^{2}+24}\right)+\frac{1}{6}\left(-a+\sqrt{a^{2}+24}\right)=0
\end{gathered}
$$

From (3) it follows that

$$
\begin{aligned}
-6 a+6 \sqrt{a^{2}+24}-a-\sqrt{a^{2}+24} & =0 \\
5 \sqrt{a^{2}+24} & =7 a \\
25 a^{2}+600 & =49 a^{2}
\end{aligned}
$$

from (4) it also follows (5). Therefore, in both cases $24 a^{2}=600, a=5$ or $a=-5$.

II. For $a=5$, (1) is $3 x^{2}+5 x-2=0$ and has as solutions the two numbers $\frac{1}{6}(-5 \pm 7)$, i.e., the numbers $\frac{1}{3}$ and -2. Both satisfy (2) as well.

For $a=-5$, (1) is $3 x^{2}-5 x-2=0$ and has as solutions the two numbers $\frac{1}{6}(5 \pm 7)$, i.e., the numbers $-\frac{1}{3}$ and 2. Both also satisfy (2).

Thus, the two numbers $a=5$ and $a=-5$ are the desired ones.

Adapted from $[5]$