$\begin{aligned} & {\left[\begin{array}{l}\text { Irrational Equations } \\ \text { [ Completing the Square. Sums of Squares }\end{array}\right]}\end{aligned}$

Solve the equation

$$
\left(x^{2}+x\right)^{2}+\sqrt{x^{2}-1}=0
$$
Since the numbers $\left(x^{2}+x\right)^{2}$ and $\sqrt{x^{2}-1}$ are non-negative, and their sum is zero, then both these numbers are equal to zero. On the other hand, if both these numbers are equal to zero, then their sum is zero. Therefore, the original equation is equivalent to the following system:

$$
\left\{\begin{array} { c } 
{ ( x ^ { 2 } + x ) ^ { 2 } = 0 ; } \\
{ \sqrt { x ^ { 2 } - 1 } = 0 }
\end{array} \Leftrightarrow \left\{\begin{array} { l } 
{ x ^ { 2 } + x = 0 ; } \\
{ x ^ { 2 } - 1 = 0 }
\end{array} \Leftrightarrow \left\{\begin{array}{l}
x \in\{0,-1\} ; \\
x \in\{1,-1\}
\end{array} \Leftrightarrow x=-1\right.\right.\right.
$$

## Answer

$x=-1$.