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Lowe put in $4000, and Taft put in $2000. When the partnership expired, they found that they had gained $9000. What was each partner's share of the gain?

7. S. Gamwell, C. Grover, R. Wheelock, and W. Godding formed a partnership, under the title of Gamwell, Grover, & Co. Gamwell at first put in $8000, but at the end of 6 mo. he withdrew $2000, and at the end of 12 mo. he withdrew $1000 more. Grover at first put in $6000, but at the end of 10 mo. he put in $3000 more. Wheelock put in $7000. Godding at first put in $10,000; at the end of 6 mo. he withdrew $2000, and at the end of 14 mo. he put in $4000. At the end of 2 years they found that they had gained $12,000. What was each man's share of the gain?

SECTION XVI.

POWERS AND ROOTS.

211. Definitions.

(a.) THE product of a number taken any number of times as a factor is called a POWER of the number. - See 105, (d.) (e.) (f.), and Note.

(b.) A ROOT of any number is such a number as, taken some number of times as a factor, will produce the given number.

(c.) If the root must be taken twice as a factor to produce the number, it is the SQUARE ROOT, or the SECOND ROOT; if three times, it is the CUBE ROOT, or the THIRD ROOT; if four times, it is the FOURTH ROOT; &c.

Thus, 2 is the square root of 4, the third root of 8, the fourth root of 16, &c., because 22 = 4, 23: 8, 24 = 16; &c.

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(d.) The character √, called the RADICAL SIGN, is used

to indicate that the root of the number over which it is placed is to be extracted.

(e.) The DEGREE of the root is indicated by a small figure, called an INDEX, which is placed a little above and at the left of the sign. When no index is written, the square root is required.

Thus,

4, or 2/4, means the square root of 4.

243 means the fifth root of 243.

73 means the fourth root of the 3d power of 7.

(f) We may also indicate that a root is to be extracted, by using a fractional exponent.

Thus, 99; (125) = 125; 27/272; &c.

(g.) The process of finding the powers of numbers is called INVOLUTION, and the process of finding their roots is called EVOLUTION, or the EXTRACTING OF Roots.

212. Relation which the Denominations of a Square bear to those of its Root.

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(b.) The above table shows that, — First. There are below 100 but 9 entire numbers which are perfect squares.

Second. The entire part of the square root of any number below 100 will be less than 10, and therefore contain but 1 figure; of any number between 100 and 10,000 will lie between 10 and 100, and therefore contain 2 figures; between 10,000 and 1,000,000 will lie between 100 and 1000, and therefore will contain 3 figures; &c.

1. How many figures are there in the entire part of the Equare root of 865698?

Answer.

Since 865698 lies between 10,000 and 1,000,000, its root

must lie between the roots of those numbers, i. e., between 100 and 1000, and must therefore contain 3 figures in its entire part.

How many figures are there in the entire part of the

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213. Division into Periods.

(a.) As the square of 10 is 100, of 100 is 10,000, &c., it follows that the square of any number of tens will be some number of hundreds; of any number of hundreds will be some number of ten thousands, &c.; or, in other words, that the square of tens will give units of no denomination below hundreds; the square of hundreds will give units of no denomination below ten-thousands; &c.

(b.) Hence, the two right hand figures of any number will contain no part of the square of the denominations of the root above units; the four right hand figures will contain no part of the square of those above tens, &c.

(c.) Therefore, if we should begin at the right of any number, and separate it into periods of two figures each, the number of periods would be the same as the number of figures in its square root. The square of the highest denomination of the root would be found in the left hand period; the square of the two highest denominations would be found in the two left hand periods; &c.

1. Separate 8478695 into periods, and explain their uses.

Answer. 8478695. The left hand period, 8, contains all of the square of the thousands of the root; the two left hand periods, 847, the square of the thousands and hundreds ; &c.

Separate each of the following numbers into periods, and explain their uses: —

2. 5794865.

3. 89475948.

4. 375486792.

5. 32500675.

214. Method of forming a Square.

(a) To find a law of universal application in squaring or extracting the square roots of numbers, we will use the letter a to represent any number whatever and b to represent any other number.

(b.) Then will a+b represent the sum, and (a + b)2, or (a + b) X (a+b) the square of the sum of any two numbers whatever.

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(c.) Performing the multiplication, we have a times a a2; a times b a X b, or, as it may be written, ab; b times a = a times b = & X b, or ab; b times b = b2.

(d.) Writing the work, as below, and adding the partial products, we have,

a+b
a+b

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(a + b) x (a + b) =a2+2 times a b + b2 = a2 + 2 a b + b2

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(e.) Hence, (a + b)2 : a2+2ab+ b2, or, since a2 equals the square of the first number, and 2 ab equals twice the product of the first number by the second, and b2 equals the square of the second;

The square of the sum of any two numbers equals the square of the first, plus twice the product of the first by the second, plus the square of the second.

b2

Illustrations.

(75)2 = 72 + 2 × 7 X 5+ 52 = 497025: =144122 (8+4)2=82 + 2 × 8 × 4 + 42: 64 +64 + 16 - 144 = €122 (20+3)2=202+2 × 3 × 20 +32=400+ 120+9=529=232

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(f.) But a2+2 ab + b2 can be put into another form; for 2 ab+ 2 a times b+b times b, = (2 a + b) times b, or (2 a + b) × b, or, by omitting the sign X, as may be done without ambiguity, (2 a + b) b.

Hence, (a+b)2 a2 + 2 a b + b2: = a2 + (2 a + b) b.

(g.) But a2 means the square of the first number; 2 a + b means the sum of twice the first number, plus the second; and (2 a + b) b the sum of twice the first, plus the second, multiplied by the second.

(h.) Hence, the square of the sum of two numbers is also equal to the square of the first number, plus the product obtained by multiplying the sum of twice the first number plus the second, by the second.

Illustrations.

(7+5)2 =72+ (14+ 5) 5

(8+4)2=82 + (16 + 4) 4

4995144 = 122.
6480144 = 122.

(408)2 = 402 + (80 +8) 8 = 1600 + 704 = 2304 = 482.

(i.) Now, as any number above ten is composed of tens and units, its square will be composed of the square of the tens, plus the product of twice the tens plus the units multiplied by the units.

(j.) If there are more than ten tens in the number, the part which is

composed of tens may be considered as made up of hundreds and tens, and its square will equal the square of the hundreds, plus the product of twice the hundreds, plus the tens, multiplied by the tens.

(k.) Proceeding in this way, we shall at last reach the part which is expressed by one or two figures, and composed of only the two highest denominations of the given number. The square of this part will be the square of the highest denomination, plus the product of twice the highest denomination, plus the next lower, multiplied by the next lower Thus,

(4837)2= (4830 +7)2:

48302 + (2 X 4830 +7) X 7

(4830)2 = (4800 + 30)2 = 48002 + (2 × 4800 + 30) × 30
(4800)2 = (4000+800)2 = 40002 + (2 × 4000 + 800) × 800

215. Method of extracting the Square Root.

What is the value of √925444 ?

Solution.(a.) Since this number lies between 10,000 and 1,000,000, its root must lie between 100 and 1000, and must therefore be composed of hundreds, tens, and units. Dividing it into periods of two figures each, it will take the form 925444.

(b.) If, now, we let a represent the hundreds of the root, and b the tens, the whole of a2 will be found in the left hand period, i. e., in the ten-thousands, and the whole of (a + b)2 in the two left hand periods, i. e., in 9254 hundreds.

(c.) The greatest square in 92 is 81, the root of which is 9. Therefore, 9 = a= the hundreds figure of the root. Subtracting a2, = 81 ten-thousands, from 92 ten-thousands, leaves 11 ten-thousands, to which adding the 54 hundreds gives 1154, which must contain (2 a + b) b.

(d.) Now, as we know a, we can find 2 a, and make use of it as a trial divisor to find b. But a being hundreds and b tens, 2 a b must be thousands, and no part of it will be found to the right of the thousands.

(e.) Hence, in dividing, we may disregard the right hand figure of 1154, and see how many times the trial divisor, 18, is contained in 115. The quotient is 6, which is probably b, the tens figure of the root. If this is correct, (2 a + b), or the true divisor, must be equal to 186, and (2 a + b) b must be equal to 6 times 186, or 1116. This last product, being less than 1154, shows that the work is correct. We subtract, and to the remainder, 38 hundreds, add the right hand period, 44 units, which gives 3844 for a new dividend.

(f.) Now, if we let a' represent the part of the root already found, i. e., the 96 tens, and b' the units, a' + b' will represent the required root, and (a' + b′)2 = a12 + (2 a' + b') b' the given number.

(g.) But we have already subtracted a/2; the remainder, 3844, must contain the (2 a + b') b'.

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