Find all values ​of $b$ for which the equation has a unique solution.

Find all values ​​of $b$ for which the equation has a unique solution.
$$x-2=\sqrt{2(b-1)x+1}$$
I found the roots of the equation. $x=b+1\pm\sqrt{b^2+2b-2}$
As I understand, there is one solution: $b=1$, then $x=3$. But I don’t understand how to find it?

Solution 1:

You need to find for which values of $b$, the equation has only ONE solution. For example, when $b=1$ you found there was only one solution which is $x=3$. For unknown $b$

$$ x-2=\sqrt{2\left(b-1\right)x+1} \Rightarrow \left(x-2\right)^2=2\left(b-1\right)x+1 $$ Hence $x$ must be a solution of $$ x^2-4x+4=2bx-2x+1\Leftrightarrow x^2-2\left(b+1\right)x+3=0 $$ Then $$ \Delta=4\left(b+1\right)^2-12=4\left(b^2+2b-2\right) $$ Can you discuss whether this kind of expression has no, one or two solutions ? What conditions on $b$ do you need on the solution $x$ so it makes sense ? (Think about the original equation containing a $\sqrt{}$.