# Task No. 3.1

## Condition:

Solve the equation $6 x^{2}+10 x+25+5 y^{2}+10 x y=0$.
# Answer:

$\mathrm{x}=-5, \mathrm{y}=5$

## Exact match of the answer -1 point

Solution.

Let's represent $6 x^{2}$ as $5 x^{2}+x^{2}$. Then the left side, if we change the order of the terms, can be represented in the following form:

$$
\left(x^{2}+10 x+25\right)+\left(5 y^{2}+10 x y+5 x^{2}\right)=0
$$

Now notice that the left side can be factored as $(x+5)^{2}+5(x+y)^{2}$.

Since $(x+5)^{2}$ is at least zero (since the square of any real number is non-negative), and $(x+y)^{2}$ is at least zero (for the same reason), and therefore $5(x+y)^{2}$ is also at least zero, then, since their sum is zero, this is only possible when both terms are zero. We get that $(x+5)^{2}=0, (x+y)^{2}=0$. If the square of a number is zero, then the number itself is also zero (because the square of non-zero numbers is greater than zero), so $x+5=0$, $x+y=0$. From this, we get that $\mathrm{x}=-5, \mathrm{y}=5$.