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.

juriemagic Feb 22, 2020

#1**+1 **

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.

---------------------------------------------------------

Here are two important facts.

\(\sqrt{9}=+3\)

The square root was in the question so the answer must be positive.

-------------

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?

Melody Feb 22, 2020

#2**+1 **

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?..

juriemagic
Feb 22, 2020