5. (8 points) Solve the equation $x^{4}-(a+b) \cdot x^{3}+(a b-2 c) \cdot x^{2}+c(a+b) \cdot x+c^{2}=0$. In your answer, write the sum of the squares of the real roots of the equation.
Solution: $x=0$ is not a solution to the equation, so we can divide by $x^{2}$. Grouping terms, we get

$$
\begin{gathered}
\left(x^{2}-2 c+\frac{c^{2}}{x^{2}}\right)-(a+b) \cdot\left(x-\frac{c}{x}\right)+a b=0 \\
\left(x-\frac{c}{x}\right)^{2}-(a+b) \cdot\left(x-\frac{c}{x}\right)+a b=0 \\
\left(x-\frac{c}{x}-a\right) \cdot\left(x-\frac{c}{x}-b\right)=0
\end{gathered}
$$

We solve the two equations $x^{2}-a x-c=0$ and $x^{2}-b x-c=0$ using the discriminant. The roots are

$$
x_{1,2}=\frac{a \pm \sqrt{a^{2}+4 c}}{2} \quad \text { and } \quad x_{3,4}=\frac{b \pm \sqrt{b^{2}+4 c}}{2}
$$

All four roots are real numbers (in all cases, the numbers were chosen such that the discriminants are non-negative). In the end, we get

$$
\begin{gathered}
\left(\frac{a-\sqrt{a^{2}+4 c}}{2}\right)^{2}+\left(\frac{a+\sqrt{a^{2}+4 c}}{2}\right)^{2}+\left(\frac{b-\sqrt{b^{2}+4 c}}{2}\right)^{2}+\left(\frac{b+\sqrt{b^{2}+4 c}}{2}\right)^{2}= \\
=\frac{a^{2}+\left(a^{2}+4 c\right)+b^{2}+\left(b^{2}+4 c\right)}{2}=a^{2}+b^{2}+4 c
\end{gathered}
$$

Unfair solution: If we believe that all four roots of the equation are real numbers, we can find the answer in a shorter way. By Vieta's theorem,

$$
\begin{gathered}
x_{1}+x_{2}+x_{3}+x_{4}=a+b \\
x_{1} x_{2}+x_{1} x_{3}+x_{1} x_{4}+x_{2} x_{3}+x_{2} x_{4}+x_{3} x_{4}=a b-2 c
\end{gathered}
$$

Therefore,

$$
\begin{gathered}
x_{1}^{2}+x_{2}^{2}+x_{3}^{2}+x_{4}^{2}=\left(x_{1}+x_{2}+x_{3}+x_{4}\right)^{2}-2\left(x_{1} x_{2}+\ldots+x_{3} x_{4}\right)= \\
=(a+b)^{2}-2(a b-2 c)=a^{2}+b^{2}+4 c
\end{gathered}
$$

## Answers:

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