4. In the set of real numbers, solve the equation

$$
\sqrt{5 x^{2}+14 x+9}-\sqrt{x^{2}-x-20}=5 \sqrt{x+1}
$$
Solution. Let's determine the domain. We have:

$$
5 x^{2}+14 x+9=5(x+1)\left(x+\frac{9}{5}\right) \geq 0, x^{2}-x-20=(x-5)(x+4) \geq 0 \text { and } x+1 \geq 0
$$

so we get that the domain is $x \geq 5$.

$$
\begin{aligned}
& \sqrt{5 x^{2}+14 x+9}-\sqrt{x^{2}-x-20}=5 \sqrt{x+1} \\
& \sqrt{(5 x+9)(x+1)}=\sqrt{(x-5)(x+4)}+5 \sqrt{x+1} \\
& 2 x^{2}-5 x+2=5 \sqrt{(x-5)(x+4)(x+1)} \\
& 2\left(x^{2}-4 x-5\right)+3 x+12=5 \sqrt{(x-5)(x+1)(x+4)} \\
& 2(x-5)(x+1)+3(x+4)=5 \sqrt{(x-5)(x+1)(x+4)} \\
& \frac{2(x-5)(x+1)+3(x+4)}{x+4}=\frac{5 \sqrt{(x-5)(x+1)(x+4)}}{x+4} \\
& \frac{2(x-5)(x+1)}{x+4}+3=5 \sqrt{\frac{(x-5)(x+1)}{x+4}}
\end{aligned}
$$

If $\sqrt{\frac{(x-5)(x+1)}{x+4}}=y$, then $2 y^{2}-5 y+3=0 \quad$ and we get $y_{1}=\frac{3}{2}, y_{2}=1$. From $\sqrt{\frac{(x-5)(x+1)}{x+4}}=\frac{3}{2}$, we sequentially get

$$
\begin{aligned}
& \frac{(x-5)(x+1)}{x+4}=\frac{9}{4} \\
& 4(x-5)(x+1)=9(x+4) \\
& 4 x^{2}-25 x-56=0
\end{aligned}
$$

and from there $x_{1 / 2}=\frac{25 \pm \sqrt{625+896}}{8}$, i.e., $x_{1}=8, x_{2}=-\frac{7}{4}$. From $\sqrt{\frac{(x-5)(x+1)}{x+4}}=1$, we sequentially get

$$
\begin{aligned}
& \frac{(x-5)(x+1)}{x+4}=1 \\
& (x-5)(x+1)=x+4 \\
& x^{2}-5 x-9=0
\end{aligned}
$$

and from there $x_{3 / 4}=\frac{5 \pm \sqrt{25+36}}{2}=\frac{5 \pm \sqrt{61}}{2}$.

Finally, the solutions to the equation are 8 and $\frac{5+\sqrt{61}}{2}$.

## III year