## Task B-2.1.

Mara has decided to reshape her rectangular flower bed into a square flower bed. If she chooses one side of the rectangular flower bed as the side of the square flower bed, the double area of the resulting square will be $12 \mathrm{~m}^{2}$ larger than the area of the rectangle. If she chooses the other side of the rectangular flower bed as the side of the square flower bed, the sum of the area of the square and twice the area of the rectangle will be $16 \mathrm{~m}^{2}$. What are the dimensions of Mara's flower bed?
## First solution.

Let $x$ and $y$ be the lengths of the sides of the rectangle. Then we have

$$
\begin{aligned}
& 2 x^{2}=x y+12 \\
& y^{2}+2 x y=16
\end{aligned}
$$

If we express $y=2 x-\frac{12}{x}$ from the first equation and substitute it into the second, we get the equation

$$
\left(2 x-\frac{12}{x}\right)^{2}+2 x\left(2 x-\frac{12}{x}\right)=16
$$

which simplifies to

$$
4 x^{2}-48+\frac{144}{x^{2}}+4 x^{2}-24-16=0
$$

After rearranging, we obtain the biquadratic equation

$$
x^{4}-11 x^{2}+18=0
$$

Then $x_{1}^{2}=9, x_{2}^{2}=2$.

If $x=\sqrt{2}$, then $y=-4 \sqrt{2}$, which is impossible. Therefore, $x=3$, and $y=2$.

The flower bed has one side of length $3 \mathrm{~m}$, and the other $2 \mathrm{~m}$.