2. a) Determine the maximum value of the expression

$$
E(x)=\frac{5 x^{2}+10 \sqrt{3} x+23}{x^{2}+2 \sqrt{3} x+4}, \quad x \in \mathbb{R}
$$

b) Determine \( x, y \in \mathbb{R} \) that satisfy the relation:

$$
\begin{array}{r}
5 x^{2}+10 \sqrt{3} x+23=\left(x^{2}+2 \sqrt{3} x+4\right)\left(2 y^{2}+4 \sqrt{6} y+20\right) . \\
\text { Prof. Negreanu Pantelimon, Alexandria }
\end{array}
$$
2.a) $E(x)=\frac{5\left(x^{2}+2 \sqrt{3} x+4\right)+3}{x^{2}+2 \sqrt{3} x+4}=5+\frac{3}{(x+\sqrt{3})^{2}+1}$

Since $(x+\sqrt{3})^{2} \geq 0$, the maximum value is achieved for $x+\sqrt{3}=0$ and is equal to 8. $2 p$

b) The relation can be written in the equivalent form

$$
\frac{5 x^{2}+10 \sqrt{3} x+23}{x^{2}+2 \sqrt{3} x+4}=2 y^{2}+4 \sqrt{6} y+20
$$

Since $\frac{5 x^{2}+10 \sqrt{3} x+23}{x^{2}+2 \sqrt{3} x+4} \leq 8$, according to point a), and $2 y^{2}+4 \sqrt{6} y+20=$ $=2\left(y^{2}+2 \sqrt{6} y+6\right)+8=2(y+\sqrt{6})^{2}+8 \geq 8, \forall y \in \mathbb{R}$, the equality will hold for $x=-\sqrt{3}$ and $y=-\sqrt{6}$