4. (7 points) Solve the equation $x^{4}-(a+b) x^{3}+(a b+2 c) x^{2}-c(a+b) x+c^{2}=0$. 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. In all cases, the numbers are chosen such that $b^{2}-4 c<0$ and $a^{2}-4 c \geqslant 0$. In the end, we find that our original equation has only two real roots

$$
x_{1,2}=\frac{a \pm \sqrt{a^{2}-4 c}}{2}
$$

Then the sum of the squares is

$$
x_{1}^{2}+x_{2}^{2}=\left(\frac{a-\sqrt{a^{2}-4 c}}{2}\right)^{2}+\left(\frac{a+\sqrt{a^{2}-4 c}}{2}\right)^{2}=a^{2}-2 c
$$

Answers:

| Option | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 |
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| Answer | 33 | 29 | 42 | 38 | 55 | 47 | 74 | 95 | 135 | 91 |