Here is a challenge for you fellow math fans out there... I am a avid math book collector. I have math books dating from the 1860's to the present day! Here is a excerpt from a math book dated 1906 on how to compute square roots by hand. Before this explanation, there is a very large section on computing square roots on polynomials. Not easy ones like (x+y)^2 but for example here is one:
Find the square root of:
20 x^2 - 22x^3 + 1 +28 x^4 +9x^6 - 8x - 12x^5
Isn't that nuts? I've never seen a problem like this in a text book that I've used. But anyways here is an explanation how to compute square roots by hand:
We then have the following rule for finding the square root of an integral perfect square.
Separate the number into periods by pointing every second digit beginning with the units' place.
Find the greatest square in the left-hand period, and write its square root as the first digit of the root; subtract the square of the first root digit from the left hand period, and to the result annex the next period.
Divide this remainder, omitting the last digit, by twice the part of the root already found, and annex the quotient to the root, and also to the trial-divisor.
Multiply the complete divisor by the root-digit last obtained and subtract the product from the remainder.
If other periods remain. proceed as before. doubling the part of the rot already found for the next trail divisor.
Could anyone explain why this procedure works? Nan, Rob, or Jing, or anyone? I am planning on studying on calculating square roots on large polynomials and be able to teach and explain. There is also a very large section in the book on cube roots of polynomials----- fun fun fun!!!!! =)
Friday, August 1, 2008
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2 comments:
We know it works, but my kids can't get it to work with this problem:
square root of 24,336
The sqrt is 156 - but - by hand, we can't get the "5" to show up in the second place of the root. We keep getting 7 instead?? 2 43 36 - first square is 1 which produces 1 on the next line -drop the second pair of 43 and the divident of 143 requires the second place of the root to be "7". Why can't we get the "5" in the second place??
Curt
Wearehislight,
Thank you for your question. I am doing the problem you've inquired about. I am getting the exact same thing. Could this explanation help you any better though, (maybe we'll figure this out together!).
Let a represent the number in accordance with the rule of 216 which is:
The square root of 100 is 10, of 10000 is 100; etc Hence the square root of a number between 1 and 100 is between 1 and 10; the square root of a number between 100 and 10000 is 10 and 100 etc. That is, the integral part of the square root of an integer of one or two digits, contains one digit; of an integer of three or four digits, contains two digits and so on.
Example 106929:
Let a represent the hundreds digit of the root, with two ciphers annexed; b the ten's digit, with one cipher annexed; and c the units' digit. Then a must be the greatest multiple of 100 whose square is less than 106929.; this we find to be 300.
Subtracting a^2 or 90000 from the given number and find the result which is 16929.
Dividing this remainder by 2a or 600 we have the quotient 28+ which suggests that b equals 20.
Adding this to 2a or 600 and multiplying the result by b or 20 we have 12400; which when subtracted from 16929 leaves 4529. Since this remainder equals (2a+2b+c)c, we can get c approximately by diving it by 2a+2b or 600+40. Dividing 4529 by 640 we have the quotient of 7 + which suggests that c =7. Adding this to 600+40 multiplying the result by 7 and subtracting the product 4529 there is no remainder. Then 300 +20 +7 or 327 is the required square root.
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