Hi Juriemagic, it is good to see you
firstly it is not right because if you substitute in x=10 into the original equation you get 6=0 which is obviously incorrect.
Read on if you want to dispute this
Also note that we are both only talking about real numbers. x is in the set of real.
When there is a square root in a question the square root value MUST be positive.
so
\(\sqrt{x-1}\) is positive
-3 is negative.
They cannot be equal. Hence there is no solution.
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Here are two important facts.
\(\sqrt{9}=+3\)
The square root was in the question so the answer must be positive.
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BUT IF
\(x^2=9\\ then\\ x=\pm\sqrt{9}=\pm3\)
The second one is different from the first one because in the second one the square root was introduced as a part of the solution.
It was not there to start with. Do you get the difference?