1. In the set of real numbers, solve the equation

$$
\sqrt{2 x-a}=x-b
$$

where $a$ and $b$ are real numbers.
The given equation is equivalent to the system

$$
2 x-a=(x-b)^{2}, x \geq b, x \geq \frac{a}{2}
$$

The equation $2 x-a=(x-b)^{2}$ can be written in the form

$$
x^{2}-2(b+1) x+a+b^{2}=0
$$

and has solutions $x_{1}=b+1-\sqrt{2 b-a+1}, x_{2}=b+1+\sqrt{2 b-a+1}$.

a) If $\frac{a}{2} > 2 b+1$, then $x_{1}$ and $x_{2}$ are not real numbers, i.e., the given equation has no real solutions.