1. In the field of real numbers, solve the equation

$$
4 x^{4}-12 x^{3}-7 x^{2}+22 x+14=0,
$$

knowing that it has four distinct real roots, and the sum of two of them is equal to the number 1.
SOLUTION. Let the sought roots be $x_{1}, x_{2}, x_{3}, x_{4}$ such that $x_{1}+x_{2}=1$. Then

$$
4 x^{4}-12 x^{3}-7 x^{2}+22 x+14=4\left(x-x_{1}\right)\left(x-x_{2}\right)\left(x-x_{3}\right)\left(x-x_{4}\right) .
$$

By comparing the coefficients of the corresponding powers of $x$, we obtain the known Viète's formulas

$$
\begin{aligned}
x_{1}+x_{2}+x_{3}+x_{4} & =3, \\
x_{1} x_{2}+x_{1} x_{3}+x_{1} x_{4}+x_{2} x_{3}+x_{2} x_{4}+x_{3} x_{4} & =-\frac{7}{4}, \\
x_{1} x_{2} x_{3}+x_{1} x_{2} x_{4}+x_{1} x_{3} x_{4}+x_{2} x_{3} x_{4} & =-\frac{11}{2}, \\
x_{1} x_{2} x_{3} x_{4} & =\frac{7}{2} .
\end{aligned}
$$

Since $x_{1}+x_{2}=1$, from (1) it follows that $x_{3}+x_{4}=2$. Equations (2) and (3) can be rewritten as

$$
\begin{aligned}
\left(x_{1}+x_{2}\right)\left(x_{3}+x_{4}\right)+x_{1} x_{2}+x_{3} x_{4} & =-\frac{7}{4} \\
\left(x_{1}+x_{2}\right) x_{3} x_{4}+\left(x_{3}+x_{4}\right) x_{1} x_{2} & =-\frac{11}{2}
\end{aligned}
$$

which, after substituting the values $x_{1}+x_{2}=1$ and $x_{3}+x_{4}=2$, gives

$$
\begin{aligned}
x_{1} x_{2}+x_{3} x_{4} & =-\frac{15}{4} \\
2 x_{1} x_{2}+x_{3} x_{4} & =-\frac{11}{2} .
\end{aligned}
$$

From this system of two linear equations, we easily obtain

$$
x_{1} x_{2}=-\frac{7}{4}, \quad x_{3} x_{4}=-2
$$

Notice that for these values of the products $x_{1} x_{2}$ and $x_{3} x_{4}$, equation (4), which we have not used so far, is also satisfied. From the conditions $x_{1}+x_{2}=1, x_{1} x_{2}=-\frac{7}{4}$, it follows that $x_{1}$ and $x_{2}$ are the roots of the quadratic equation

$$
x^{2}-x-\frac{7}{4}=0, \quad \text { i.e., } x_{1,2}=\frac{1}{2} \pm \sqrt{2}
$$

Similarly, from the conditions $x_{3}+x_{4}=2$ and $x_{3} x_{4}=-2$, we get

$$
x_{3,4}=1 \pm \sqrt{3}
$$

There is no need to check the found roots, as we have emphasized, the entire system of equations (1) to (4) is satisfied.

Conclusion. The given equation has roots $\frac{1}{2}+\sqrt{2}, \frac{1}{2}-\sqrt{2}, 1+\sqrt{3}, 1-\sqrt{3}$.