4. Solve the equation $\sqrt{5-x}=x^{2}-5$.

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Solution. The domain of the equation is the set $( -\infty,-\sqrt{5}) \cup$ $(\sqrt{5}, 5)$. For $x \in(-\infty,-\sqrt{5}) \cup(\sqrt{5}, 5)$, the given equation is equivalent to the equation $5-x=\left(x^{2}-5\right)^{2}$. We will write the equation as a quadratic in the number 5, i.e., $5^{2}-\left(1+2 x^{2}\right) 5+x^{4}+x=0$. From there, we obtain the equations:

$$
5=\frac{\left(1+2 x^{2}\right)+(1-2 x)}{2} \Leftrightarrow x^{2}-x-4=0 \text { and } 5=\frac{\left(1+2 x^{2}\right)-(1-2 x)}{2} \Leftrightarrow x^{2}+x-5=0
$$

By direct verification, we observe that only $x_{1}=\frac{-1-\sqrt{21}}{2}, x_{2}=\frac{1+\sqrt{17}}{2}$ belong to the domain.

## III year