#### Alan1

I am looking at two different methods of solving one simple problem, that give conflicting answers, and would like to know which of my solutions is wrong.

The problem is simple: simplify the square root of a square. Example:

(x2)(1/2)
Here are steps that prove the answer is x:

(x2)(1/2) = x(1/2) * x(1/2) = x(1/2 + 1/2) = x(2/2) = x1 = x

However, consider the case where x = -4. By the above process:

((-4)2)(1/2) = (-4)(1/2) * (-4)(1/2) = (-4)(1/2 + 1/2) = (-4)(2/2) = (-4)1 = -4

However, I could also solve it this way:

((-4)2)(1/2) = 16(1/2) = 4

One way, I get an answer of -4, the other way, I get an answer of 4, and every step is correct for both solutions.

Can someone please resolve this for me? Where did I do something that is incorrect?

#### skeeter

MHF Helper
Attached is a graph of $y= \sqrt{x^2}$. What does it tell you?

#### Archie

You should also graph $$\displaystyle y=\left(\sqrt{x}\right)^2$$.

It's also vitally important that you are aware of what number system you are working with. In particular, if you are in the real numbers, then the $$\displaystyle \sqrt{-4}$$ in the original post does not exist. If you are in the complex numbers, then the square root function has two branches.

#### Shakarri

Both 4 and -4 squared give 16 so if you take the square root of 16 you can get either one. There are two valid answers for the square root of 16.

#### Plato

MHF Helper
I am looking at two different methods of solving one simple problem, that give conflicting answers, and would like to know which of my solutions is wrong.
The problem is simple: simplify the square root of a square. Example:
(x2)(1/2)
Here are steps that prove the answer is x:
(x2)(1/2) = x(1/2) * x(1/2)
Although post #2 has shown you that is incorrect, I will add a post script.
It is quite common for students to mix meanings. That is: square-root four and $\bf{\sqrt{4}}$ are really different things.
The first refers to two values $\pm 2$ whereas $\sqrt{4}=2$ (only one positive number).
As Skeeter said, $\sqrt{x}$ exists if and only if $x\ge 0$ and it is always is true that $\sqrt{x}\ge 0$.