4. (7 points) Solve the equation

$$
x+\sqrt{x+\frac{1}{2}+\sqrt{x+\frac{1}{4}}}=n^{2} \quad n>1
$$
Solution: Let's complete the square in the expression under the root

$$
x+\sqrt{\left(\sqrt{x+\frac{1}{4}}+\frac{1}{2}\right)^{2}}=n^{2}
$$

The root can be removed without the absolute value due to the fact that in the domain of definition (ODZ) $\sqrt{x+\frac{1}{4}}+\frac{1}{2}>0$. Therefore,

$$
x+\sqrt{x+\frac{1}{4}}+\frac{1}{2}=n^{2}
$$

Again, we complete the square

$$
\left(\sqrt{x+\frac{1}{4}}+\frac{1}{2}\right)^{2}=n^{2}
$$

Since $n>0$ and in the ODZ $\sqrt{x+\frac{1}{4}}+\frac{1}{2}>0$, then

$$
\sqrt{x+\frac{1}{4}}=n-\frac{1}{2}
$$

Moreover, we know that $n>1$, so squaring will be an equivalent transformation

$$
x+\frac{1}{4}=n^{2}-n+\frac{1}{4} \Longrightarrow x=n^{2}-n
$$

## Answers:

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