Task B-2.2. One solution of the equation $a x^{2}+b x+5=0$ is five times larger than the other. The coefficients $a$ and $b$ are natural numbers less than 20. Determine all such equations.
## Solution.

From the given condition $x_{1}=5 x_{2}$ and Vieta's formulas

$$
\begin{aligned}
x_{1}+x_{2} & =-\frac{b}{a} \\
x_{1} \cdot x_{2} & =\frac{c}{a}
\end{aligned}
$$

it follows that

$$
\begin{aligned}
6 x_{2} & =-\frac{b}{a} \\
5 x_{2}^{2} & =\frac{5}{a}
\end{aligned}
$$

If we express $x_{2}$ from the first equation and substitute it into the second, we get

$$
\left(-\frac{b}{6 a}\right)^{2}=\frac{1}{a}
$$

i.e., $b^{2}=36 a$.

Since $a$ and $b$ are natural numbers, $a$ can only be the square of some natural number. Given that $a$ and $b$ are less than 20, it follows that $a=1, b=6$ or $a=4, b=12$ or $a=9, b=18$.

(1 point)

The solutions to the equations are

$x^{2}+6 x+5=0$, $4 x^{2}+12 x+5=0$ and $9 x^{2}+18 x+5=0$.

(1 point)