
1. Find all real roots of the equation

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

if it is known that it has four distinct real roots, two of which sum up to 1.


Solution. Denote the roots by $x_{1}, x_{2}, x_{3}, x_{4}$ in such a way that $x_{1}+x_{2}=1$. Then

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

Comparing the coefficients at the corresponding powers of $x$, we obtain the familiar Vièta's relations

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

Since $x_{1}+x_{2}=1$, it follows from (1) that $x_{3}+x_{4}=2$. We rewrite the equations (2) and (3) in the form

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

Upon substituting $x_{1}+x_{2}=1$ and $x_{3}+x_{4}=2$, this yields

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

From this system of linear equations it is already easy to obtain

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

Observe that for these values of the products $x_{1} x_{2}$ and $x_{3} x_{4}$, the 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

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

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

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

Since, as we have already remarked, these roots satisfy all equations (1) to (4), they are also a solution to the original problem.

Conclusion. The roots of the equation are $\frac{1}{2}+\sqrt{2}, \frac{1}{2}-\sqrt{2}, 1+\sqrt{3}$, and $1-\sqrt{3}$.

Other solution. From the hypothesis it follows that the left-hand side of the equation is the product of polynomials

$$
x^{2}-x+p \quad \text { and } \quad 4 x^{2}+q x+r
$$

where $p, q$ and $r$ are real numbers. Upon multiplying out and comparing coefficients at the corresponding powers of $x$, we obtain a system of four equations with three unknowns

$$
\begin{aligned}
q-4 & =-12, \\
4 p-q+r & =-7 \\
p q-r & =22 \\
p r & =14 .
\end{aligned}
$$

The first three equations have a unique solution $r=-8, p=-\frac{7}{4}$ and $q=-8$, which also fulfills the fourth equation. Thus we arrive at the decomposition

$$
4 x^{4}-12 x^{3}-7 x^{2}+22 x+14=\left(x^{2}-x-\frac{7}{4}\right)\left(4 x^{2}-8 x-8\right) \text {. }
$$

The equation $x^{2}-x-\frac{7}{4}=0$ has roots $\frac{1}{2} \pm \sqrt{2}$, and the equation $4 x^{2}-8 x-8=0$ has roots $1 \pm \sqrt{3}$.
