# Task 2. 15 points

Solve the equation

$$
\sqrt{\frac{x^{3}+5}{1+\sqrt{5}}}=x
$$

#
# Solution

The equation is equivalent to the system

$$
\sqrt{\frac{x^{3}+5}{1+\sqrt{5}}}=x \Leftrightarrow\left\{\begin{array} { c } 
{ \frac { x ^ { 3 } + 5 } { 1 + \sqrt { 5 } } = x ^ { 2 } } \\
{ x \geq 0 , }
\end{array} \Leftrightarrow \left\{\begin{array} { c } 
{ x ^ { 3 } + 5 = x ^ { 2 } ( 1 + \sqrt { 5 } ) , } \\
{ x \geq 0 }
\end{array} \Leftrightarrow \left\{\begin{array}{c}
5-\sqrt{5} x^{2}+x^{3}-x^{2}=0 \\
x \geq 0
\end{array}\right.\right.\right.
$$

In the obtained equation, let $a=\sqrt{5}$, leading to a quadratic equation in terms of $a$:

$$
a^{2}-x^{2} a+x^{3}-x^{2}=0
$$

Solving this equation for $a$, we get:

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

From which:

$$
\left[\begin{array} { c } 
{ x ^ { 2 } - x - \sqrt { 5 } } \\
{ x = \sqrt { 5 } }
\end{array} \Leftrightarrow \left[\begin{array}{c}
x=\frac{1 \pm \sqrt{1+4 \sqrt{5}}}{2} \\
x=\sqrt{5}
\end{array}\right.\right.
$$

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

Answer: $\frac{1+\sqrt{1+4 \sqrt{5}}}{2}, \sqrt{5}$.

## Criteria: