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### Square Root of a Negative Number Squared

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Date: 01/23/97 at 12:05:48
From: Cindy Smith
Subject: Re:  Exponent Rules

Hi,

I'm reading popular math books in addition to my textbook and am
really trying to understand the material.  Here's my question:

I understand that the nth root of x^n = x^(n/n) or x^1.  My textbook
gives two problems:  Sqrt(-6)^2 and the 4th root of (-3)^4.  This
leads to two different solutions in my mind.  (-6)^2 = 36 so the
square root of (-6)^2 = 6. Similarly, the 4th root of (-3)^4 leads to
the fourth root of 81, which is 3. However, according to the rules,
the nth root of x^n = x^(n/n). Therefore, the square root of (-6)^2
should equal (-6)^(2/2) or (-6)^1, which equals -6.  By a similar
method, the fourth root of (-3)^4 = (-3)^(4/4) or (-3)^1, which
equals -3.  The answers in the back of the book says the correct
answers are positive 6 and 3.  Where is the error in my logic?

Thanks again for all your help!

Cindy Smith
```

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Date: 01/23/97 at 20:28:04
From: Doctor Keith
Subject: Re:  Exponent Rules

Great question!  Please note that all questions are welcome, so you
should write as many as you like.  We are here to help!

About the root question, both you and the book are right in a way.
Your book gives the positive number because, by convention, you always
take the positive and real root.  Thus Sqrt((-3)^2) = 3.

It is important to note that while this is the primary root it is by
no means the only one.  In fact you can't get a positive root for
problems like:

((-1)^3)^(1/3)

Note that the key to this problem is raising a negative number to an
odd power so it stays negative. There is clearly no positive number
that you can cube to get a negative number. I mention this as a
warning on the positive solution because your book is careful to pick
even powers so they do not have to deal with this problem.

I would also like to commend you on noticing that the square root
operation yields two roots.  You might be interested to know that if
we consider complex numbers (ordered pairs of real numbers, which can
be pictured as a plane with the real numbers on the x axis and the
imaginary numbers, from sqrt(-1), on the y axis), and you take the nth
root (ie x^(1/n)), you will actually get n answers! So taking cube
roots in the complex plane yields 3 distinct answers, each of which
when cubed yields the original number. It provides a beautiful
symmetry, and is extremely useful in many applications (I am an
engineer and I use this literally all the time).  I find this really
neat, but if you have not seen complex numbers before, fear not, you
will see them all in good time.  Something to look forward too.

To sum up, the bottom line is your book has probably used the
definition that the positive real root is the answer (this is
sometimes called the primary root) when it exists.  For instance, the
square root of a positive number is a positive number (this is always
possible).  Note that the square root of a negative number would not
have an answer using this definition, and it is for this case that the
are just not what your book wanted.

If you would like to see a physical interpretation of your answers do
the following:

1) for Sqrt((-6)^2) write: x = Sqrt((-6)^2)

2) square both sides: x^2 = (-6)^2

3) then we have: x^2 - (-6)^2 = 0

4) graph y = x^2 - 36 and see where it crosses the x-axis. You will
see it crosses at both 6 and -6.

The graphing trick works because when the graph of y = x^2 - 36
crosses the x-axis we have that y = 0 so we have 0 = x^2 -36. This
will reinforce the idea that both roots are valid. The reason we
sometimes ignore the negative root is due to the system we are working
with. Frequently we only need the positive root (in many systems we
must have positive numbers such as when calculating the thickness of
some material we need, since thickness measures of a part must be
positive).  We sometimes call these system requirements by the name
constraints.  One other reason that your book deals with positive
roots is the simple fact that they are nicer to work with.  For
instance, if we consider a fourth root as a square root of a square
root, we have to deal with square roots of negative numbers, which
will involve complex numbers (sorry, no way around it).

Hope I didn't give you too much.  If something is confusing reply back
as to the problem and I will try to give you some more information or
a book reference if you prefer.  Again, you should be very proud of
your observation.  Hope this helps.  Good luck.

-Doctor Keith,  The Math Forum
Check out our web site!  http://mathforum.org/dr.math/
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Associated Topics:
High School Imaginary/Complex Numbers
High School Negative Numbers
High School Square & Cube Roots
Middle School Negative Numbers
Middle School Square Roots

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