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Sqrt (x-2) = Sqrt (x+2) ?

Date: 01/10/2003 at 15:29:24
From: Ace
Subject: Math Puzzle

Is the following equation solvable for x?

   sqrt (x-2) = sqrt (x+2)

Does it have something to do with imaginary numbers?

Any help is appreciated!


Date: 01/10/2003 at 20:28:22
From: Doctor Jeremiah
Subject: Re: Math Puzzle

Hi Ace,

Squaring both sides is exactly what must be done:

       sqrt(x-2) = sqrt(x+2)
             x-2 = x+2
               x = x+2+2
               x = x+4

So the question is equivalent to asking "what number is equal to 
itself plus 4?" And of course we could continue this algebra like 
this:

       sqrt(x-2) = sqrt(x+2)
             x-2 = x+2
               x = x+2+2
               x = x+4
             x-x = 4
               0 = 4

The fact that we made zero equal to four means that there is NO answer 
to the original question. It can't be solved for x because sqrt(x-2) 
can NEVER be equal to sqrt(x+2).

Whenever you get one number equal to another, that means either you 
made an algebra mistake or the original equation is invalid (which is 
the case this time).

If it were imaginary numbers then x would have the form of x = a + 
sqrt(-1)b and if we substitute that in we get:

       sqrt(x-2) = sqrt(x+2)
             x-2 = x+2
 a+sqrt(-1)b - 2 = a+sqrt(-1)b + 2
     a+sqrt(-1)b = a+sqrt(-1)b + 4
   a + sqrt(-1)b = a+4 + sqrt(-1)b

And we are back in the same place: The real part must be equal to 
itself plus 4, and we can again simplify the whole thing to 0 = 4, 
which means the original equation is not valid.

So, in the end, all we can say for sure is that the square root of 
(x-2) can NEVER be equal to the square root of (x+2)

- Doctor Jeremiah, The Math Forum
  http://mathforum.org/dr.math/ 
Associated Topics:
High School Basic Algebra
High School Square & Cube Roots

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