+0

# square root equals negative

0
67
3
+603

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.

Feb 22, 2020

#1
+108679
+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?

Feb 22, 2020
#2
+603
+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
#3
+108679
+1

I absolutely agree.

You should always test or think about whether an answer is correct or reasonable.

Sometimes in exams, there may not be much time for checking but even under exam conditions, a quick check should be done.

Melody  Feb 22, 2020