Find the root of the following equation, with a precision of 3 significant digits.

$$
(\sqrt{5}-\sqrt{2})(1+x)=(\sqrt{6}-\sqrt{3})(1-x)
$$
By multiplication and rearrangement

$$
x=\frac{(\sqrt{6}+\sqrt{2})-(\sqrt{5}+\sqrt{3})}{(\sqrt{6}-\sqrt{2})+(\sqrt{5}-\sqrt{3})}
$$

and the denominator is clearly positive. We multiply the numerator and the denominator by the difference of the two bracketed expressions in the denominator:

$$
x=\frac{4-2 \sqrt{30}+2 \sqrt{6}+2}{(8-2 \sqrt{12})-(8-2 \sqrt{15})}=\frac{3-\sqrt{30}+\sqrt{6}}{\sqrt{15}-\sqrt{12}}=\frac{\sqrt{3}-\sqrt{10}+\sqrt{2}}{\sqrt{5}-2}
$$

then further by $\sqrt{5}+2$:

$$
x=\sqrt{15}-3 \sqrt{2}-\sqrt{10}+2 \sqrt{3}=\sqrt{15}+\sqrt{12}-(\sqrt{18}+\sqrt{10})
$$

To calculate the decimal form of $x$ to the required accuracy, we need to know the place value of the first significant digit of $x$. According to the school four-digit function table, rounded to the thousandth place,

$$
\sqrt{15} \approx 3.873, \quad \sqrt{12} \approx 3.464, \quad \sqrt{18} \approx 4.243, \quad \sqrt{10} \approx 3.162
$$

from these, $x \approx -0.068$. The error of each of the four roundings is less than 0.0005, so the error of $x$ is less than $4 \cdot 0.0005 = 0.002$, therefore the place value of the first significant digit of $x$ is 1/100. Thus, we need to determine the 1/1000 and 1/10000 place values of $x$, in other words, we need to calculate $x$ rounded to four decimal places.

We will show that to achieve the required accuracy (in this example), it is sufficient to determine the 5-digit lower approximations of the square roots in (1). According to the known method of square root extraction,

$$
\begin{array}{ll}
\sqrt{15}=3.87298 \ldots, & \sqrt{12}=3.46410 \ldots \\
\sqrt{18}=4.24264 \ldots, & \sqrt{10}=3.16227 \ldots
\end{array}
$$

thus the following inequalities hold for the terms of (2):

$$
\left\{\begin{aligned}
3.87298 & <\sqrt{15}<3.87299 \\
3.46410 & <\sqrt{12}<3.46411 \\
-4.24265 & <-\sqrt{18}<-4.24264 \\
-3.16228 & <-\sqrt{10}<-3.16227
\end{aligned}\right.
$$

By adding these,

$$
-0.06785<x<-0.06781
$$

The rounded value to four decimal places of any number satisfying this double inequality is $-0.0678$, so this is the value of $x$ we are looking for.

Tegze Judit (Budapest, Kölcsey F. g. I. o. t.)

Remark. We transform the expression for $x$ obtained in (1) into a product. Then, "rationalizing the numerator" and multiplying again:

$$
\begin{gathered}
x=-(\sqrt{6}-\sqrt{5})(\sqrt{3}-\sqrt{2})=-\frac{1}{(\sqrt{6}+\sqrt{5})(\sqrt{3}+\sqrt{2})}= \\
=-\frac{1}{\sqrt{18}+\sqrt{15}+\sqrt{12}+\sqrt{10}}
\end{gathered}
$$

Here, by reducing or increasing the data in (2) by 0.0005, similar to (3), we find that the denominator falls between 14.740 and 14.744, and $x$ is the negative reciprocal of these: between $-0.067843$ and $-0.067824$ (the first rounded down, the second up), so to three significant figures, $x=-0.0678$.