Problem 1. a) Determine the real numbers a, b, c knowing that $a^{2}+b^{2}+c^{2}=14$ and $2a+4b+6c=28$.

b) Let $t=(\sqrt{2}+\sqrt{3}) \cdot(a \sqrt{2}+b \sqrt{3})$, where a and b are integers. If $t$ is a rational number, calculate $x=a|b|+b|a|$.
Solution:

![](https://cdn.mathpix.com/cropped/2024_06_07_e3d15057ad6f2c4f2a36g-1.jpg?height=62&width=1515&top_left_y=1080&top_left_x=305)

The relation above can be written as: $\left(a^{2}-2 a+1\right)+\left(b^{2}-4 b+4\right)+\left(c^{2}-6 c+9\right)=0$. 1 point

This results in: $(a-1)^{2}+(b-2)^{2}+(c-3)^{2}=0$ 1 point

Since each term of the sum is non-negative $\Rightarrow a-1=b-2=c-3=0$.

0.5 points

Thus: $a=1, b=2, c=3$.

0.5 points

b) The given number can be written as: $t=(2 a+3 b)+\sqrt{6}(a+b)$

0.5 points

This results in: $\sqrt{6}(a+b)=t-(2 a+3 b)$

0.5 points

Since the right-hand side of the equality is a rational number $\Rightarrow \sqrt{6}(a+b) \in \mathbb{Q}$

0.5 points

Since $(a+b) \in \mathbb{Z}$ and $\sqrt{6} \in \mathbb{R}-\mathbb{Q} \Rightarrow a+b=0 \Rightarrow b=-a$

1 point

We obtain $x=a|-a|-a|a|=0$