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Square Roots using Basic Arithmatic


Date: 03/19/2002 at 19:55:52
From: Tony Starks
Subject: Square Roots using basic arithmatic

There are three values needed. Two are given: X and Y. Solve for Z.

X = Z^Y.

Example X = Z^Y, X = 729, Y = 6
        729 = Z*Z*Z*Z*Z*Z

How can I solve for Z using only addition, subtraction, 
multiplication, and division?


Date: 03/19/2002 at 22:32:34
From: Doctor Ian
Subject: Re: Square Roots using basic arithmatic

Hi Tony,

The easiest way would be to find the prime factors of 729:

  729 = 3 * 243

      = 3 * 3 * 81

      = 3 * 3 * 9 * 9

      = 3 * 3 * 3 * 3 * 3 * 3

Does this help? 

- Doctor Ian, The Math Forum
  http://mathforum.org/dr.math/   


Date: 03/19/2002 at 23:30:22
From: Tony Starks
Subject: Square Roots using basic arithmatic

Yes, but how would I do that for any value given to X?


Date: 03/20/2002 at 10:00:22
From: Doctor Ian
Subject: Re: Square Roots using basic arithmatic

Hi Tony,

Finding the prime factors of a number is straightforward, although 
tedious: 

   Prime Factorization
   http://mathforum.org/dr.math/problems/primefac.html   

It's possible that if you're given a series of problems, it will be 
possible to solve them all by using this method. (For example, the 
point of having you solve the problems might be to give you practice 
in recognizing and playing with powers of integers.) 

Sometimes the method may be able to get you to a simpler problem, 
e.g., 

  
   X = Z^Y, X = 9604, Y = 4
    
   9604 = Z*Z*Z*Z

The prime factors of 9604 are

  9604 = 2 * 2 * 7 * 7 * 7 * 7

With a little finagling, we can break them up this way:

  9604 = sqrt(2)^4 * 7^4

       = (sqrt(2) * 7)^4

which means that X = 7*sqrt(2). 

Now, if you get a problem like 

   X = Z^Y, X = 214.39, Y = 5
    
   614.39 = Z*Z*Z*Z*Z

then you have to treat it as an ordinary root problem, which you 
answer by guessing.

Seriously. You would play with some integers to get close:

  2^5 =  32         Too small

  3^5 = 243         Too small

  4^5 = 1024        Too large

So now you know it's somewhere between 3 and 4.  Is it 3.5?

  3.5^5 = 525.22

Still too small. So now you know it's in the interval (3.5,4.0). 

And you go from there. This requires nothing more than multiplication 
to do, but it requires a lot of multiplication, and it is guaranteed 
to get you as close to the answer as you want.  

I guess the main thing I'd like to get across to you is that you 
always want to look for ways to apply tricks like prime factorization 
first, because when they work, they work quickly:

   Why study Prime and Composite Numbers?
   http://mathforum.org/dr.math/problems/howell.01.25.01.html   
  
And _only_ when you run out of tricks do you fall back on general 
methods.    

Let me give you another example of what I'm talking about. Suppose 
you're asked to solve two equations:

  2x + 3y = 16
  
  2x + 5y = 20

Now, there are several general methods (substitution, elimination, and 
so on) that you can use to attack this problem, by manipulating the 
equations in various ways. Or, you could pause to look for a way to 
_see_ the answer. In this case, we can notice that in going from the 
first equation to the second, we've added two extra y's on the left, 
and 4 extra whatevers on the right. Which means that 

  2y = 4

which makes finding the value of x trivial. It's _only_ when you can't 
find an easy way to proceed that you resort to the 'usual methods'.  

Does this help?  Write back if you'd like to talk more about this, or 
anything else. 

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

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