A Way to Estimate the Square Root of Any Number
Date: 03/16/2006 at 13:00:42 From: JP Subject: Wow! I have found it, now why does it work? Today in math class, we learned about finding the square root of certain numbers, such as the square root of 36 = 6. I found a way to find the approximate square root of any number, and I need to know why it works. Example: the square root of 63. To get the square root of 63 first you take the nearest two perfect squares, 49 and 64. Take the square roots of those numbers, which are 7 and 8. Set it up like this: 49 63 64 7 ? 8 Next we need to make a fraction. Subtract 49 from 64 to get your denominator, 15. To get your numerator you subtract the first two numbers, 63 - 49 = 14. That leaves you with 14 over 15 or .93. Since the square root of 63 is higher than 7 but less than 8 your whole number is 7. Add .93 to 7 and get 7.93, close to the real square root of 63. In my math class we always question why. So, why does this work?
Date: 03/16/2006 at 15:03:47 From: Doctor Peterson Subject: Re: Wow! I have found it, now why does it work? Hi, JP. What you've discovered is called "linear interpolation". It amounts to approximating a point on a graph by using a straight line that is close to the graph. Draw the graph of the square root, y = sqrt(x). It passes through the points (49,7) and (64,8), since the square roots of 49 and 64 are 7 and 8 respectively. Between them, the graph is slightly curved. If you draw a line between those two points, you'll find that the line is quite close to the graph of the square root. Here's a close- up of the line: 8 -------------------- o / | o | / | |1 / |h | / | | 7 -- o-----------+-----+ | | | 49 63 64 \_________/ 63-49=14 \_______________/ 64-49=15 Since 63 is 14/15 of the way from 49 to 64, h is 14/15 of the way from 7 to 8. That means the value of y, when x is 63, is 7 14/15. That's your estimate of the square root. This method is often used for quick estimates; back before calculators, we were taught to use it to find values from trigonometric tables when we needed extra accuracy. Here's a picture of the full graph that I talked about in case you'd like to see it, with the triangle I drew in red. You can see that the straight line is very close to the actual square root graph: Did you discover this on your own, or did you have some hints that led you in this direction? It's a very nice thing to have found! If you have any further questions, feel free to write back. - Doctor Peterson, The Math Forum http://mathforum.org/dr.math/
Date: 03/16/2006 at 21:26:41 From: JP Subject: Wow! I have found it, now why does it work? I discovered it on my own and my teacher was amazed at the discovery. She had never heard of figuring it out in that way. Thanks for your reply. That makes sense. Jacob
Search the Dr. Math Library:
Ask Dr. MathTM
© 1994- The Math Forum at NCTM. All rights reserved.