6.128. Find all values of $a$ for which the sum of the roots of the equation $x^{2}-2 a(x-1)-1=0$ is equal to the sum of the squares of the roots.
## Solution.

Given $x^{2}-2 a x+(2 a-1)=0$. By Vieta's theorem $\left\{\begin{array}{l}x_{1}+x_{2}=2 a, \\ x_{1} \cdot x_{2}=2 a-1 .\end{array}\right.$ Further,

$$
x_{1}+x_{2}=x_{1}^{2}+x_{2}^{2}=\left(x_{1}+x_{2}\right)^{2}-2 x_{1} x_{2} \Leftrightarrow x_{1}+x_{2}=\left(x_{1}+x_{2}\right)^{2}-2 x_{1} x_{2} .
$$

Using the values $x_{1}+x_{2}=2 a$ and $x_{1} \cdot x_{2}=2 a-1$, we get $2 a=(2 a)^{2}-2(2 a-1)$, which simplifies to $2 a^{2}-3 a+1=0, a_{1}=\frac{1}{2}, a_{2}=1$.

Answer: $a_{1}=\frac{1}{2}, a_{2}=1$.