+1124 Hi friends, I have a problem understanding this sum. according to the text book this sum has no solution, I do not understand why, because:
\(\sqrt{x-1}+3=0\)
\(\sqrt{x-1}=-3\)
\(x-1=(-3)^2\)
\(x-1=9\)
\(x=10\)
why would this be incorrect?. Thank you all for helping out. 
+118608 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?
+1124 Hello Melody,
Thank you sooo much for your esponse.I do see my error. my problem is as with sooo many other sums, that I tend to just do the sum, like I did in my example, and do not really comprehend, or see the restrictions. Also, I hardly ever test the result back in the original equation, to test if the answer does indeed make sense or not...something i should really start learning to do...you agree?..
