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floor of which the carpeting will cover? that is, what is one side of a square, which contains 625 square yards ? →
We have seen, (T 35,) that the contents of a square surface is found by multiplying the length of one side into itself, that is, by raising it to the second power; and hence, having the contents (625) given, we must extract its square root to find one side of the room.
This we must do by a sort of trial: and,
1st. We will endeavour to ascertain how many figures there will be in the root. This we can easily do, by pointing off the number, from units, into periods of two figures each; for the square of any root always contains just twice as many, or one figure less than twice as many figures, as are in the root; of which truth the pupil may easily satisfy himself by trial. Pointing off the number, we find, that the root will consist of two figures, a ten and a unit.
2d. We will now seek for the first figure, that is, for the tens of the root, and it is plain, that we must extract it from the left hand period 6, (hundreds.) The greatest square in 6 (hundreds) we find, by trial, to be 4, (hundreds,) the root of which is 2, (tens, 20;) therefore, we set 2 (tens) in the root. The root, it will be recollected, is one side of a square. Let us, then, form a square, (A, Fig. I.) each side of which shall be supposed 2 tens,= 20 yards, expressed by the root now obtained.
The contents of this square are 20 × 20 400 yards, now disposed of, and which, consequently, are to be deducted from the whole number of yards, (625,) leaving 225 yards. This deduction is most readily performed by subtracting the square number 4, (hundreds,) or the square of 2, (the figure in the root already found,) from the period 6, (hundreds,) and bringing down the next period by the side of the remainder, making 225, as before.
3d. The square A is now to be enlarged by the addition of the 225 remaining yards; and, in order that the figure may retain its square form, it is evident, the addition must be made on two sides. Now, if the 225 yards be divided by the length of the two sides, (20+20=40,) the quotient will be the breadth of this new addition of 225 yards to the sides c d and b c of the square A.
But our root already found, 2 tens, is the length of one side of the figure A; we therefore take double this root, = 4 tens, for a divisor.
The divisor, 4, (tens,) is in reality 40, and we are to seek how many times 40 is contained in 225, or, which is the same thing, we may seek how many times 4 (tens) is contained in 22, (tens,) rejecting the right hand figure of the dividend, because we have rejected the cipher in the divisor. We find our quotient, that is, the breadth of the addition, to be 5 yards; but, if we look at Fig. II., we shall perceive that this addition of 5 yards.to the two sides does not complete the square; for there is still wanting, in the corner D, a small square, each side of
which is equal to this last quotient, 5; we must, therefore, add this quotient, 5, to the divisor, 40, that is, place it at the right hand of the 4, (tens,) making it 45; and then the whole divisor, 45, multiplied by the quotient, 5, will give the contents of the whole addition around the sides of the figure A, which, in this case, being 225 yards, the same as our dividend, we have no remainder, and the work is done. Consequently, Fig. II. represents the floor of a square room, 25 S*
yards on a side, which 625 square yards of carpeting will exactly cover.
The proof may be seen by adding together the several parts of the figure, thus :~
The square A contains 400 yards.
The figure B
From this example and illustration we derive the following general
Or we may prove it by involution, thus:25 × 25 = 625, as before.
FOR THE EXTRACTION OF THE SQUARE ROOT.
I. Point off the given number into periods of two figures each, by putting a dot over the units, another over the hundreds, and so on. These dots show the number of figures of which the root will consist.
II. Find the greatest square number in the left hand period, and write its root as a quotient in division. Subtract the square number from the left hand period, and to the remainder bring down the next period for a dividend.
III. Double the root already found for a divisor; seek how many times the divisor is contained in the dividend, excepting the right hand figure, and place the result in the root, and also at the right hand of the divisor; multiply the divisor, thus augmented, by the last figure of the root, and subtract the product from the dividend; to the remainder bring down the next period for a new dividend.
IV. Double the root already found for a new divisor, and continue the operation as before, until all the periods are brought down.
Note 1. If we double the right hand figure of the last divisor, we shall have the double of the root.
Note 2. As the value of figures, whether integers or decimals, is determined by their distance from the place of units, so we must always begin at unit's place to point off the given number, and, if it be a mixed number, we must point it off both ways from units, and if there be a deficiency in any period of decimals, it may be supplied by a cipher. It is plain, the root must always consist of so many integers
and decimals as there are periods belonging to each in the given number.
EXAMPLES FOR PRACTICE.
2. What is the square root of 10342656 ?
10342656 (3216, Ans.
3. What is the square root of 43264 ?
43264 ( 208, Ans.
4. What is the square root of 998001 ?
Ais. 999 Ans. 15'3.
7. What is the square root of '001296 ?
Ans. '036. Ans. '54. Ans. 6031. Ans. 12'8+.
T108. In this last example, as there was a remainder, after bringing down all the figures, we continued the operation to decimals, by annexing two ciphers for a new period, and thus we may continue the operation to any assigned degree of exactness; but the pupil will readily perceive, that he can never, in this manner, obtain the precise root; for the last figure in each dividend will always be a cipher, and the
last figure in each divisor is the same as the last quotient figure; but no one of the nine digits, multiplied into itself, produces a number ending with a cipher; therefore, whatever be the quotient figure, there will still be a remainder. Ans. 1473+. Ans. 3'16+.
11. What is the square root of 3?
Note. We have seen, (¶ 105, ex. 9,) that fractions are squared by squaring both the numerator and the denominator. Hence it follows, that the square root of a fraction is found by extracting the root of the numerator and of the denominator. The root of 4 is 2, and the root of 9 is 3. Ans..
15. What is the square root of?
When the numerator and denominator are not exact squares, the fraction may be reduced to a decimal, and the approximate root found, as directed above.
19. What is the square root of
SUPPLEMENT TO THE SQUARE ROOT.
1. What is involution? 2. What is understood by a » power? 3. the first, the second, the third, the fourth power? 4. What is the index, or exponent? 5. How do you involve a number to any required power? 6. What is evolution? 7. What is a root? 8. Can the precise root of all numbers be found? 9. What is a surd number? 10. -8 rational? 11. What is it to extract the square root of any number? 12. Why is the given sum pointed into periods of two figures each? 13. Why do we double the root for a divisor? 14. Why do we, in dividing, reject the right hand figure of the dividend? 15. Why do we place the quotient figure to the right hand of the divisor? 16. How may we