# Thread: Finding solution of x

1. ## Finding solution of x

Here is the problem: x^(2) ^(1/2) (That's the square root of x squared) = -x consists of ?

A. zero only
B. nonpositive real numbers only
C. positive real numbers only
d. all real numbers
e. no real numbers

Explanation: Choice B is correct. Every positive number n hast two square roots, one positive and the other negative, but the square root of n denotes the positive number whose square is n.

Why does it only denote the positive number? I am used to puttingi + or - whenever I take the square root of something. I just want to understand this clearly. If I wanted to simplify it more, could I also do that to solve? Thanks.

2. ## Re: Finding solution of x

I hate this with a livid passion myself, but in function theory, in order to define something as a function, it has to 'take' one starting value and 'generate' one ending value. Taking the square root is "one-to-many" - it takes one value, and generates two, and therefore needs to be restricted in order to be classed as a function.

Other examples include $\displaystyle \sqrt[3]{a+ib}$ and $\displaystyle arcsin(x)$

Many say that $\displaystyle \sqrt{x^2}=|x|$ which is the easiest way to "play the mediator" as it were.

3. ## Re: Finding solution of x

Originally Posted by benny92000
Here is the problem: x^(2) ^(1/2) (That's the square root of x squared) = -x consists of ?

A. zero only
B. nonpositive real numbers only
C. positive real numbers only
d. all real numbers
e. no real numbers

Explanation: Choice B is correct. Every positive number n hast two square roots, one positive and the other negative, but the square root of n denotes the positive number whose square is n.

Why does it only denote the positive number? I am used to puttingi + or - whenever I take the square root of something. I just want to understand this clearly. If I wanted to simplify it more, could I also do that to solve? Thanks.
You say "I am used to putting + or - whenever I take the square root of something." For example, if the problem were to solve "$\displaystyle x^2= a$" you would write "$\displaystyle x= \pm\sqrt{a}$". And you would write that precisely because "$\displaystyle \sqrt{a}$" does NOT include both.

4. ## Re: Finding solution of x

That makes more sense now. Thanks!