8. Specify the greatest value of the parameter $a$ for which there is a unique solution to the system

$$
\left\{\begin{array}{l}
y=1-\sqrt{x} \\
a-2(a-y)^{2}=\sqrt{x}
\end{array}\right.
$$
# Solution:

![](https://cdn.mathpix.com/cropped/2024_05_06_3205747dae2ad15e28c2g-04.jpg?height=162&width=1787&top_left_y=1024&top_left_x=183)
Equation: $\left\{\begin{array}{l}\sqrt{x}=1-y \\ a-2(a-y)^{2}=1-y\end{array}\right.$, considering that $1-y \geq 0$, we solve the resulting quadratic equation:

$$
\begin{aligned}
& a-2(a-y)^{2}=1-y \quad \Rightarrow \quad 2 y^{2}-y(4 a+1)+2 a^{2}-a+1=0 \\
& y_{1,2}=\frac{(4 a+1) \pm \sqrt{(4 a+1)^{2}-8\left(2 a^{2}-a+1\right)}}{4}=\frac{4 a+1 \pm \sqrt{16 a-7}}{4}
\end{aligned}
$$

For the equation to have a unique solution, and thus for the original system to have a unique solution, either the discriminant must be zero and the root must be no greater than one, or one of the two roots satisfies the condition while the other does not.

![](https://cdn.mathpix.com/cropped/2024_05_06_3205747dae2ad15e28c2g-04.jpg?height=605&width=1644&top_left_y=1816&top_left_x=189)

Therefore, the maximum value of the parameter for which the system has a unique solution is $a=2$.

Answer: 2.