# Task 2. 15 points

Solve the equation

$$
\sqrt{\frac{x^{3}+7}{1+\sqrt{7}}}=x
$$

#
# Solution:

The equation is equivalent to the system

$$
\begin{aligned}
& \qquad \sqrt{\frac{x^{3}+7}{1+\sqrt{7}}}=x \Leftrightarrow\left\{\begin{array} { c } 
{ \frac { x ^ { 3 } + 7 } { 1 + \sqrt { 7 } } = x ^ { 2 } , } \\
{ x \geq 0 , }
\end{array} \Leftrightarrow \left\{\begin{array} { c } 
{ x ^ { 3 } + 7 = x ^ { 2 } ( 1 + \sqrt { 7 } ) , } \\
{ x \geq 0 , }
\end{array} \Leftrightarrow \left\{\begin{array}{c}
7-\sqrt{7} x^{2}+x^{3}-x^{2}=0, \\
x \geq 0,
\end{array}\right.\right.\right. \\
& \text { In the obtained equation, let } a=\sqrt{7}, \text { we come to a quadratic equation } \\
& \text { in terms of } a \text { : } \\
& \qquad a^{2}-x^{2} a+x^{3}-x^{2}=0 .
\end{aligned}
$$

Solving this equation in terms of $a$, we get:

$$
\left[\begin{array}{c}
a=x^{2}-x \\
a=x
\end{array}\right.
$$

From which:

$$
\left[\begin{array} { c } 
{ x ^ { 2 } - x - \sqrt { 7 } } \\
{ x = \sqrt { 7 } }
\end{array} \Leftrightarrow \left[\begin{array}{c}
x=\frac{1 \pm \sqrt{1+4 \sqrt{7}}}{2} \\
x=\sqrt{7}
\end{array}\right.\right.
$$

Considering the condition $x \geq 0$, we get

Answer: $\frac{1+\sqrt{1+4 \sqrt{7}}}{2}, \sqrt{7}$.

## Criteria