4.66 Find all real roots of the equation
$$x^{4}-\left(2 \cdot 10^{10}+1\right) x^{2}-x+10^{20}+10^{10}-1=0$$

accurate to four decimal places.
[Solution] By Descartes' Rule of Signs:
If the coefficient sequence of the real-coefficient $n$-degree algebraic equation
$$a_{0} x^{n}+a_{1} x^{n-1}+a_{2} x^{n-2}+\cdots+a_{n-1} x+a_{n}=0\left(a_{0} \neq 0, a_{n} \neq 0\right)$$

is
$$\left\{a_{0}, a_{1}, a_{2}, \cdots, a_{n-1}, a_{n}\right\}$$

and the number of sign changes in this sequence is $p$, then the number of positive roots of the equation (counting a $k$-fold root as $k$ roots) is equal to $p$ or less than $p$ by a positive even number.

Given that the number of sign changes in the coefficient sequence of the known equation is 2, it follows that the known equation has at most two positive roots.

Transform the known equation into
$$\left(x^{2}-10^{10}-\frac{1}{2}\right)^{2}=x+\frac{5}{4} \text {. }$$

First, we prove that equation (1) has no negative roots. If it did, suppose the equation (1) has a negative root $x_0, f\left(10^{5}\right)0$ Therefore, one of the two positive roots of the equation lies between $10^{5}-1$ and $10^{5}$, and the other lies between $10^{5}$ and $10^{5}+1$.

Let $x=10^{5}+c$, where $0 < c < 1$. Then
$$\begin{aligned}
f\left(10^{5}+0.0016\right) & =\left(10^{5}+0.0016\right)^{2}-10^{10}-0.5-\sqrt{10^{5}+0.0016+1.25} \\
& =10^{10}+0.0032 \times 10^{5}+0.0016^{2}-10^{10}-0.5-\sqrt{10^{5}+1.25016} \\
& =0.0032 \times 10^{5}+0.0016^{2}-0.5-\sqrt{10^{5}+1.25016} \\
& <0
\end{aligned}$$
and
$$\begin{aligned}
f\left(10^{5}+0.0015\right) & =\left(10^{5}+0.0015\right)^{2}-10^{10}-0.5-\sqrt{10^{5}+0.0015+1.25} \\
& =10^{10}+0.003 \times 10^{5}+0.0015^{2}-10^{10}-0.5-\sqrt{10^{5}+1.2515} \\
& =0.003 \times 10^{5}+0.0015^{2}-0.5-\sqrt{10^{5}+1.2515} \\
& >0
\end{aligned}$$

Therefore, one root of the equation lies in $\left(10^{5}+0.00155,10^{5}+0.00165\right)$ and the other in $\left(10^{5}-0.00165,10^{5}-0.00155\right)$, so the two approximate roots of the equation are $x \approx 10^{5} \pm 0.0016$.