Example 3. Find the real roots of the equation
$$
\mathrm{f}(\mathrm{x})=\mathrm{x}^{41}+\mathrm{x}^{3}+1=0
$$
It is easy to prove that $f(\mathrm{x})$ is strictly monotonically increasing, and it is an odd polynomial, therefore, it has a unique real root. When $x \geqslant 0$, $f(x) \geqslant 1$, hence the original equation has no positive root. Let $x=-s$, (10) can be transformed into:
$$
g(s)=s^{4}+s^{3}-1=0 .
$$

Since $g(0)=-1, g(1)=1$, there is a root in $(0,1)$. Also, $g(0.9)=-0.2577$, and $g(0.7)$ and $g(1)$ have opposite signs, so there is a root in $(0.9, 1)$. Take $\mathrm{s}_{0}=0.95$.
$$
g^{\prime}(\dot{\varepsilon}, 0)=41 \mathrm{~s}^{40}+3 \mathrm{~s}^{2} \quad,=0.05=7.97 \mathrm{~b} \approx 8 \text { (approx. }
$$
If the approximate formula $\frac{g\left(s_{0}+h\right)-g\left(s_{0}-h\right)}{2 h}$ is used, taking $\mathrm{h} \cdots 0.01$, the same result can be obtained. Thus, the following iterative format can be established:
$$
\mathrm{s}_{\mathrm{s}+1}=\mathrm{s}_{\mathrm{n}}+\frac{1}{-8}\left(\mathrm{~s}_{\mathrm{n}}{ }^{41}+\mathrm{s}_{\mathrm{n}}{ }^{3}-1\right) .
$$

Let $x_{0}=0.95$, and repeatedly press the following keys: $\overline{S T O}$
$$
\begin{array}{l}
\left.\overline{y^{2} \mid} 41\right]\left[\overline{[(\mid}\left|\overline{M_{c}}\right| \quad\left|\overline{y^{\mathbf{x}} \mid} \quad 3\right| \bar{I}\right. \\
1 \text { I } 8 \sqrt{+1-1}+\mathbf{M}_{c}|=| \text { to get: } \\
\mathrm{s}_{1}=0.9525673 \text {. } \\
\mathrm{s}_{2}=0.9524778 \text {, } \\
s_{3}=0.9524843 \text {, } \\
s_{4}=s_{5}=0.9524838 \text {. } \\
\end{array}
$$

Thus, the real root of equation (10) is $x=-0.9524838$, with an absolute error $<10^{-7}$. In the SS-53B calculator, STO indicates storing the displayed content into memory, $\bar{M}_{\mathrm{c}}$ indicates displaying the content of the memory, which remains unchanged after display. When using the calculator, it is essential to make full use of its storage function.

Additionally, the iterative method and the simplified Newton's method can also be used to find complex roots.

Higher-order equations and transcendental equations are both types of nonlinear equations, and there are many more methods for solving such equations. Students who are interested can refer to [4].
(Author's affiliation: Tianjin Normal University, Department of Mathematics)