Drexel dragonThe Math ForumDonate to the Math Forum

Ask Dr. Math - Questions and Answers from our Archives
_____________________________________________
Associated Topics || Dr. Math Home || Search Dr. Math
_____________________________________________

Equation without a Solution


Date: 11/14/2001 at 18:25:37
From: Steven and Richard
Subject: Sqrt(x) = -2, order of operations, and other problems with 
the square root

What is the solution to the equation sqrt(x) = -2 ?  

Some people say that there is no solution, since the square root is 
defined to be the side of a square with a given area. We allow for 
negative area by allowing these imaginary numbers. If I were to have a 
negative side the area would most certainly be imaginary, or would it?  

We thought of looking at sqrt(4i^4). Using the laws of exponents 
first, the answer is most definitely -2. Using the laws of imaginary 
numbers first, the answer is 2. Is there an order of operations for 
which law to use first?

Where is the breakdown in the square root? I have ceased calling 
the square root a function because 4 = 4i^4 and y = sqrt(x) seems to 
be providing me with two real solutions. If we allow negative area, 
don't we have to allow negative sides?


Date: 11/14/2001 at 22:40:10
From: Doctor Peterson
Subject: Re: Sqrt(x) = -2, order of operations, and other problems 
with the square root

Hi, Steven.

When you start thinking about negative or imaginary numbers, it's best 
to think algebraically rather than geometrically. A square root of a 
number is defined as any number whose square is that number. A 
positive number has two square roots, one positive and the other 
negative. When we write the radical sign (which we're representing as 
"sqrt"), that represents only the POSITIVE square root of the number; 
so the two square roots of 4 are called sqrt(4) and -sqrt(4). We do 
this so that the square root is a function with one value, not 
something that somehow takes two values at once.

Therefore, your equation

    sqrt(x) = -2

has no solutions, simply because by definition sqrt(x) is always 
positive. 

This has nothing to do with imaginary numbers, which are the solutions 
of equations like

    x^2 = -2

Although these look similar, they have no real solutions for entirely 
different reasons.

Going back to geometry, if a square could have sides with negative 
lengths (which it can't), then the area would still be positive, since 
the product of two negative numbers is positive. Only with imaginary 
sides (?) could a square have a negative area.

Now, when you move into complex numbers, there is NO square root 
FUNCTION. You're exactly right: any definition that chooses a single 
square root for every complex number will be inconsistent, in the 
sense that

    sqrt(ab) = sqrt(a) sqrt(b)

will not always be true. It's only a fortunate convenience that a 
square root function over the reals can be defined. This means that 
you can't look to the complex numbers for a solution to your equation, 
because any equation involving square roots must allow only reals.

You will be interested in these pages that deal with various issues 
you have raised:

   Square Root of 100 - Dr. Math archives
   http://mathforum.org/dr.math/problems/cry.04.15.99.html   

   What is i? - Dr. Math archives
   http://mathforum.org/dr.math/problems/toeti24.html   

   False Proofs, Classic Fallacies - Dr. Math FAQ
   http://mathforum.org/dr.math/faq/faq.false.proof.html   

(see the link near the bottom to "1 = 2: A Proof using Complex 
Numbers").

- Doctor Peterson, The Math Forum
  http://mathforum.org/dr.math/   
    
Associated Topics:
High School Exponents
High School Number Theory

Search the Dr. Math Library:


Find items containing (put spaces between keywords):
 
Click only once for faster results:

[ Choose "whole words" when searching for a word like age.]

all keywords, in any order at least one, that exact phrase
parts of words whole words

Submit your own question to Dr. Math

[Privacy Policy] [Terms of Use]

_____________________________________
Math Forum Home || Math Library || Quick Reference || Math Forum Search
_____________________________________

Ask Dr. MathTM
© 1994-2013 The Math Forum
http://mathforum.org/dr.math/