Problem 4. Solve the inequality

$$
\sqrt[4]{x^{2}+2 x+10}+x^{4} \leqslant 62-32 x
$$
Solution. Since $x^{2}+2 x+10=(x+1)^{2}+9>0$ for any value of $x$, the given inequality is defined for any real value of $x$. Let's investigate some properties of the continuous function

$$
f(x)=\sqrt[4]{(x+1)^{2}+9}+x^{4}+32 x-62
$$

If $x \geqslant 0$, then the continuous function $f(x)$ is increasing. Therefore, the equation $f(x)=0$ has no more than one non-negative root $x_{1}$. But $f(0)0$, so $00$ is true for any value of $x$. Therefore, the equation $f(x)=0$ does not have roots for those values of $x$ that are solutions of the inequality $x^{4}+32 x-62 \geqslant 0$.

We get $x^{4}+32 x-62=x\left(x^{3}+32\right)-62$. If $x \leqslant-4$, then $x^{3}+32128$. Therefore, for $x \leqslant-4$, the inequality $x^{4}+32 x-62>0$ holds. Now it is clear that the negative roots of the equation $f(x)=0$ can only belong to the interval $(-4 ; 0)$.

Using a programmable microcalculator, we construct a table of values of the function $f(x)$:

| $x$ | -4 | $-3.5$ | -3 | $-2.5$ | -2 | $-1.5$ | -1 | $-0.5$ | 0 |
| :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: |
| $f(x)$ | 68 | -22 | -75 | -101 | -108 | -103 | -91 | -76 | -60 |

Based on this table, we can assume that the equation $f(x)=0$ has no roots on the interval $[-3.5 ; 0]$. But how do we prove this?

Obviously,

$$
\sqrt[4]{9} \leqslant \sqrt[4]{(x+1)^{2}+9} \leqslant \sqrt[4]{(-3.5+1)^{2}+9}
$$

Hence,

$$
1.73 \leqslant \sqrt[4]{(x+1)^{2}+9} \leqslant 1.98
$$

Let's determine how the function $\varphi(x)=x^{4}+$ $+32 x-62$ changes on the interval $[-3.5 ; 0]$. We find $\varphi^{\prime}(x)=$
$=4 x^{3}+32$. Solving the equation $4 x^{3}+32=0$, we get $x=-2$. We calculate:

$\varphi(-2)=-110 ; \varphi(0)=-62 ; \varphi(-3.5) \approx-23.9375$. (4)

From conditions (3) and (4), it is clear that on the entire interval $[-3.5 ; 0]$, the function $f(x)0$, and $f(-3.5) \approx-22<0$, so there exists a unique root $x_{2}$ of the equation $f(x)=0$ on the interval $(-4 ;-3.5)$.

To refine the values of the roots $x_{1}$ and $x_{2}$, we construct a table of values of the function $f(x)$ using a microcalculator:

| $x$ | $f(x)$ | $x$ | $f(x)$ |
| :--- | :---: | :---: | :---: |
|  |  |  |  |
| 0 | $-60.22$ | -4 | 68.06 |
| 2 | 20.06 | -3 | $-75.1$ |
| 1.5 | $-7.00$ | $-3.5$ | $-21.96$ |
| 1.6 | $-2.25$ | $-3.7$ | 9.02 |
| 1.7 | 2.76 | $-3.65$ | 0.6897 |
| 1.65 | 0.9754 | -35 | $-0.92905$ |
| 1.64 | $-0.287$ | $-3.645$ | $-0.12166$ |
| 1.643 | $-0.13743$ | $-3.646$ | 0.04029 |
| 1.644 | $-0.087508$ |  |  |
| 1.645 | $-0.037553$ |  |  |
| 1.646 | 0.012434 |  |  |

Thus, $x_{1} \approx-3.6455 ; x_{2} \approx 1.6455$.

It is easy to notice that $x_{1}+x_{1} \approx-2, x_{1} x_{2} \approx-6$. Solving the system of equations: $x_{1}+x_{2}=-2, x_{1} x_{2}=-6$, we get the hypothesis:

$$
x_{1}=-\sqrt{7}-1, x_{2}=\sqrt{7}-1
$$

By direct substitution of $x_{1}$ and $x_{2}$ into the equation $f(x)=0$, we confirm the validity of this hypothesis.