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Computing Square Roots Manually

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Date: 03/05/98 at 23:23:36
From: Jerome Ross
Subject: Square Roots

Hello, my name is Jerome Ross and I am a student at the University of
Phoenix. I am currently taking a college algebra class. I was given an
assignment to learn how to compute square roots manually. If you could
help me, I would greatly appreciate it.

Thank You,

Jerome Ross
```

```
Date: 03/08/98 at 14:38:50
From: Doctor Sam
Subject: Re: Square Roots

Jerome,

There is an old technique that is similar to long division, but I
can't explain it through this medium. You might check any pre-1950
algebra textbook.

But there is another method, the bisection method, that I think I can
explain here. To get the idea of this method, consider the problem of
factoring a number like 12

12 = 1 x 12 = 2 x 6 = 3 x 4

or a number like 64, which factors into

64 = 1 x 64 = 2 x 32 = 4 x 16 = 8 x 8.

In general, when a number N = AB, then when A = B, you have found the
square root of N. When A > B, then the square root must lie between A
and B.

Let's apply this idea to finding the square root of 500 numerically.
Start by factoring 500 ... numbers that are more nearly equal will get
you to the square root faster than 1 x 500, although starting there

So 500 = 50 x 10. We know that 10 < sqrt(500) < 50. We also know that
the average of 10 and 50 is between 10 and 50, as well:

(10+50)/2 = 30.

Divide 500 by 30, resulting in 50/3, so that 500 = 30 x (50/3). Since

50/3 < 30

it must be true that

50/3 < sqrt(500) < 30.

We started with sqrt(500) between 10 and 50 and have now narrowed the
range to (50/3, 30).

If we keep repeating the method, we can get sqrt(500) to any desired
accuracy. Here are several iterations of the method:

500 = 10 x 50         Aver.(10,50)       = 30
Divide 500/30      = 50/3

500 = (50/3)x 30      Aver.(50/3,30)     = 70/3
Divide 500/(70/3)  = 150/7

500 = (150/7)x(70/3)  Aver.(150/7, 70/3) = 22.381 approx.

500 = (22.381)(22.340)

Notice that after four repetitions of the process, we have the two
factors equal to the tenths place. Repeated applications will continue
to halve the interval separating the factors (that's why it is called
the bisection method).

I hope that helps.

-Doctor Sam, The Math Forum
Check out our web site http://mathforum.org/dr.math/
```
Associated Topics:
High School Square & Cube Roots
Middle School Square Roots

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