Re: An old algorithm for Square Roots
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Re: An old algorithm for Square Roots
They dont teach this in the algebra books these days. Sometimes, you dont
even get it until calculus. Teachers haev become used to the square root button.
At 11:29 PM 08-09-98 -0500, GARY WARDALL wrote:
>The following is an excerpt from ALGEBRA IN EASY STEPS Third Edition by
>Edwin I. Stein D. of Van Nostrand Company 1956 p243. This is the algorithm
>that was required in our school district up until about 1970. We learned it,
>in most cases including me, not too well. We certainly did know why it worked
>or if it would always work. This is a good example of knowing how to
>use an algorithm but not understanding the concepts behind it. By the way
>the reason it works is based on the fact that one solution of (A+x)^2=B for
>small
>x relative to A can be approximated by the solution to the much easier equation
>A^2+2*a*x=B.
>
>Good Luck
>Gary Wardall
>*********************************
>
>EXERCISE 93 Square Roots
>
>I. Aim: To find the square root of a number.
>
>II. Procedure
>
>1. Separate the number into groups of two figures each, starting at the
>decimal point
> and forming the groups, first to the left and then to the right of the
>decimal point.
> Note: If there is an odd number of figures to the left of the decimal
>point, there will
> be one group containing a single number. However if there is an odd
>number of figures
> to the right of the decimal point, add a zero so that each group
>contains two figures.
>
>2. Find the largest square which can be subtracted from the first group at
>the left.
> Write it under the first group.
>
>3. Write the square root of this largest square above the first group as
>the first figure
> of the square root.
>
>4. Subtract the square number from the first group. Annex the next group to
>the remainder.
>
>5. Form the trial divisor by multiplying the root already found by 2 and
>annexing a zero.
> Note:In the sample solutions the zero is not written but is used mentally.
>
>6. Divide the remainder (step 4) by the trial divisor (step 5). Annex the
>quotient to the root
> already found; also add it to the trial divisor to form the complete
divisor.
>
>7. Multiply the complete divisor by the new figure of the root.
>
>8. Subtract this product (step 7) from the remainder (step 4).
>
>9. Continue this process until all the groups, have been used or the
>desired number of decimal
> places has been obtained.
>
>10. Since each figure of the root is placed directly above its
>corresponding group, the decimal
> point in the root is placed directly above the decimal point in the
>given number.
>
>11. Check by squaring the root to obtain the given number.
>
>III. Sample Solutions
>
>1. Find the square root of 328,329.
>
> 5 7 3
> \/ 3 2 8 3 2 9
> 2 5
> ___
> 107)7 8 3
> 7 4 9
> _____
> 1143) 3 4 2 9
> 3 4 2 9
> _______
> ....
>
> Answer, 573
>
>
>2. Find the square root of 935.2 correct to nearest hundredth.
>
>
> 3 0. 5 8
> \/9 3 5. 2 0 0 0
> 9
> ___
> 605)3 5 2 0
> 3 0 2 5
> _______
> 6108) 4 9 5 0 0
> 4 8 8 6 4
> _________
> 6 3 6
>
>
>
>Answer, 30.58
>
>**********************************************
>
>
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