## Task B-2.1.

Determine all real numbers $k$ for which one solution of the equation

$$
(x-2 k)^{2}=k(x+2)
$$

is 1 greater than the other.
## Solution.

Given the equation, after squaring and rearranging, can be written in the following form:

$$
\begin{array}{r}
x^{2}-4 x k+4 k^{2}-k x-2 k=0 \\
x^{2}-5 k x+4 k^{2}-2 k=0
\end{array}
$$

By applying Vieta's formulas, we get

$$
x_{1}+x_{2}=-\frac{b}{a}=5 k, x_{1} \cdot x_{2}=\frac{c}{a}=4 k^{2}-2 k \text {. }
$$

According to the condition in the problem, $x_{1}=x_{2}+1$, so after substituting into Vieta's formula for the sum of the solutions, we have

$$
x_{2}+1+x_{2}=5 k
$$

or

$$
x_{2}=\frac{5 k-1}{2}, x_{1}=\frac{5 k+1}{2}
$$

Substitute the obtained values into Vieta's formula for the product of the solutions. Then we have

$$
\frac{5 k-1}{2} \cdot \frac{5 k+1}{2}=4 k^{2}-2 k
$$

or

$$
\begin{aligned}
25 k^{2}-1 & =16 k^{2}-8 k, \\
9 k^{2}+8 k-1 & =0
\end{aligned}
$$

The solutions to this equation, the numbers $k_{1}=-1$ and $k_{2}=\frac{1}{9}$, are the sought real numbers.