6. (8 points) Solve the equation

$$
\sqrt{a^{2}+c-1-2 x-x^{2}}+\sqrt{b^{2}+c-1-2 x-x^{2}}=\sqrt{x^{2}+2 x+1-c+(a+b)^{2}}
$$

In your answer, write the sum of the squares of all possible pairwise differences of the real roots of the equation. Numbers $a, b>0$.
Solution: Let's transform our equation and make the substitution $t=x+1$:

$$
\sqrt{a^{2}+c-t^{2}}+\sqrt{b^{2}+c-t^{2}}=\sqrt{t^{2}-c+(a+b)^{2}}
$$

Notice that if $t$ is a root of the equation, then $(-t)$ is also a root. Therefore, it is sufficient to find all non-negative roots. From now on, we will assume that $t \geqslant 0$.

As $t$ increases, the functions $\sqrt{a^{2}+c-t^{2}}$ and $\sqrt{b^{2}+c-t^{2}}$ decrease, while $\sqrt{t^{2}-c+(a+b)^{2}}$ increases. Therefore, there is no more than one solution to the given equation. Notice that $t=\sqrt{c}$ is a solution to the equation. Therefore, the original equation has exactly 2 roots: $x_{1,2}= \pm \sqrt{c}-1$. Let's find the answer to the problem:

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

## Answers:

| Option | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 |
| :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: |
| Answer | 52 | 72 | 76 | 80 | 84 | 56 | 44 | 92 | 60 | 68 |