Longhand Square RootsDate: 03/30/98 at 11:48:43 From: Tim Subject: Square root How do you find the square root of any number? Is there an easy formula? Date: 03/30/98 at 15:33:57 From: Doctor Rob Subject: Re: Square root There are a couple of ways. See http://mathforum.org/dr.math/faq/faq.sqrt.by.hand.html for one. Here is another: To find a square root by the "longhand" method, we proceed as follows. I intersperse numbered steps with an example. We will find the square root of 113 to three decimal places. 1. Draw a square root symbol, or radical, with the number whose root you are seeking underneath. Start with the decimal point and mark off digits in both directions in groups of two. Put a decimal point above the radical, and directly above the other decimal point. . /------------- \/ 1 13.00 00 00 2. Start with the first group of 1 or 2 digits. Find the largest square of a single-digit integer less than or equal to it. Write the single digit above the radical, and its square under the first group. Draw a line under that square, and subtract it from the first group. 1 . /------------- \/ 1 13.00 00 00 1 ---- 0 3. Bring down the next group below the last line drawn. This forms the current remainder. Draw a vertical line to the left of the resulting number, and to the left of that line put twenty times the number above the radical, a plus sign, a blank space (to be filled in during step 4), an equals sign, and some blank space for the answer. 1 . /------------- \/ 1 13.00 00 00 1 ---- 20+_=?? | 0 13 4. Pick the biggest digit D that you could put into the underscore place, so that when you do the indicated addition and then multiply the sum by D, the product is less than the current remainder. (If you guess too large a D, the remainder will be negative. If you guess too small a D, the remainder will be greater than the number to the left of the vertical line.) Put it above the radical above the last group of digits brought down, and also put it in the blank space you left in step 3. Compute the number given by the expression, and put it after the equals sign. Multiply D times that number, and put that below the current remainder, draw a horizontal line below that, and subtract, to give a new current remainder. 1 0. /------------- \/ 1 13.00 00 00 1 ---- 20+0=20 | 0 13 0 ----- 13 5. If the current answer, above the radical, has the desired accuracy, stop. Otherwise, go back to step 3. Step 3: 1 0. /------------- \/ 1 13.00 00 00 1 ---- 20+0=20 | 0 13 0 ----- 200+_=??? | 13 00 Step 4: 1 0. 6 /------------- \/ 1 13.00 00 00 1 ---- 20+0=20 | 0 13 0 ----- 200+6=206 | 13 00 12 36 --------- 64 Step 3: 1 0. 6 /------------- \/ 1 13.00 00 00 1 ---- 20+0=20 | 0 13 0 ----- 200+6=206 | 13 00 12 36 --------- 2120+_=???? | 64 00 Step 4: 1 0. 6 3 /------------- \/ 1 13.00 00 00 1 ---- 20+0=20 | 0 13 0 ----- 200+6=206 | 13 00 12 36 -------- 2120+3=2123 | 64 00 63 69 -------- 31 Step 3: 1 0. 6 3 /------------- \/ 1 13.00 00 00 1 ---- 20+0=20 | 0 13 0 ----- 200+6=206 | 13 00 12 36 -------- 2120+3=2123 | 64 00 63 69 -------- 21260+_=????? | 31 00 Step 4: 1 0. 6 3 0 /------------- \/ 1 13.00 00 00 1 ---- 20+0=20 | 0 13 0 ----- 200+6=206 | 13 00 12 36 --------- 2120+3=2123 | 64 00 63 69 -------- 21260+0=21260 | 31 00 0 ----- 31 00 Step 5: Stop. Thus the square root of 113 to three decimal places is 10.630. Checking, 10.630^2 = 112.9969, and 10.631^2 = 113.0182, so the answer is correct. The underlying principle is (10*a+b)^2 = 100*a^2 + b*(20*a+b), or even better, (100*N+n) - (10*a+b)^2 = (100*[N-a^2]+n) - b*(20*a+b). Here is a more detailed explanation of what is going on: N is the integer made up of first groups of digits, and a is the integer part of the square root of N already calculated. N - a^2 is the current integer remainder. When you bring down the next two-digit group n, and append it to the integer remainder, you are multiplying the integer remainder by 100 and adding n to give 100*[N-a^2] + n. The 20*a represents doubling the previous answer and adding a decimal place, and the "+b" and the "b*" are the new digit you are adding and multiplying by. The old square root was a, and the new one is 10*a + b, gotten by appending the digit b to the old square root. Then (100*N+n) - (10*a+b)^2 is the new integer remainder. b is chosen so that this remainder is positive, but as small as possible. That ensures that at each step, b will be just a single digit. Now replace N by 100*N+n, and a by 10*a+b, and repeat. -Doctor Rob, The Math Forum http://mathforum.org/dr.math/ |
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