Contradictory answers for nested exponents (ex: ((-4)^2)^(1/2)

Feb 2016
1
0
USA
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
Jun 2008
16,217
6,765
North Texas
Attached is a graph of $y= \sqrt{x^2}$. What does it tell you?
 

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Dec 2013
2,002
757
Colombia
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.
 
Oct 2012
751
212
Ireland
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
Aug 2006
22,507
8,664
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 $\sqrt{x}\ge 0$.