Problem 17. Calculate the smallest root $x_{0}$ of the equation

$$
x^{2}-\sqrt{\lg x+100}=0
$$

(with a relative error of no more than $10^{-390} \%$ ).
Solution. Equation (1) is defined on $[10^{-100}; +\infty)$. Let $P(x) = x^2 - \sqrt{\lg x + 100}$. Obviously, $P(10^{-100}) > 0; P(1) < 0; P(0.1) < 0$.

Replace equation (1) with the equivalent equation $x^4 = \lg x + 100$. Denote: $M(x) = x^4, \quad F(x) = \lg x + 100$. We find $M'(x) = 4x^3, F'(x) = 1/(x \ln 10)$.

If $10^{-100} < x < 0.1$, then $M'(x) < F'(x)$. From this, it follows that the smallest root of the given equation belongs to the interval $(10^{-100}; 10^{-1})$. Other roots of equation (1) do not belong to this interval.

Replace equation (1) with the equivalent equation

$$
x = \sqrt[4]{\lg x + 100}
$$

If $x = 10^{-100 + 10^{-338}}$, then $\sqrt[4]{\lg x + 100} = 10^{-99}$. Since $10^{-100 + 10^{-396}} < 10^{-99}$, the smallest root of equation (1) belongs to the interval $(10^{-100}; 10^{-100 + 10^{-396}})$. But

$$
\frac{10^{-100 + 10^{-396}} - 10^{-100}}{10^{-100}} = 10^{10^{-396}} - 1
$$

Now it remains to prove that $10^{10^{-336}} < 10^{-392}$. Using a microcalculator, we get $\sqrt[10^6]{10} < 1.000003$. Since $\sqrt[n]{1 + a} < 1 + \frac{a}{n}$ (if $0 < a < 1$), then

$$
10^{10^{-336}} = \sqrt[10^{330}]{\sqrt[100^6]{10}} < \sqrt[10^n]{1.000003} < 1 + 10^{-395}
$$

Answer: $10^{-100} < x_0 < 10^{-100 + 10^{-396}}$.