Computing Square Roots Manually
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 will get an answer eventually. 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/
Search the Dr. Math Library:
Ask Dr. MathTM
© 1994- The Math Forum at NCTM. All rights reserved.