(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. (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 = 8 X b, or a b; b times b = b2. (d.) Writing the work, as below, and adding the partial products, we have, a2+2 times a b + b2 = a2 + 2 a b + b2 = (e.) Hence, (a + b)2 a2+2 a b + 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. Illustrations. (7 + 5)2 = 72 + 2 × 7 × 5 + 52 = 49 + 70 + 25 = 144 = 122 (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 (f.) But a2 + 2 ab + b2 can be put into another form; for 2 ab + b2 = 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 = 49 + 95 = 144 = 122. = 144122. = 1600704 2304482. (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 (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, e., in 9254 hundreds. i. (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 + b2)2 = a/2 + (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'. (h.) We now make 2 a', or 192, a trial divisor; and since 2 a/ b' must be tens, we omit the right hand figure of 3844, and see how many times 2 a', or 192, is contained in 384. (i.) This gives b' = 2, = the probable units figure. If it be correct, the true divisor, 2 a' + b, must equal 1922, and (2 a' + b') b' must be 2 X 1922. This, being equal to 3844, shows that the given number is a perfect square, and 962 is its root. (j.) If there had been another figure in the root, we might have represented the part of the root already found by a", or by some other letter, and the next figure by b", or by some other letter, and have proceeded as before. (k.) The numerical work would be written thus: 216. Rule for Square Root, with Problems. As a similar process can always be followed, we may describe the method of extracting the square root of a number thus : — First. Divide the given number into periods of two figures, beginning with the units. Second. Find the greatest square in the left hand period, and place its root as the highest denomination of the required root. Third. Subtract the square thus found from the left hand period, and to the remainder bring down the next period, calling the result a dividend. Fourth. Double the part of the root already found for a trial divisor. Fifth. See how many times this trial divisor is contained in all of the dividend, excepting the right hand figure, and write the quotient as the next figure of the root, and also place it at the right of the trial divisor, to form a true divisor. Sixth. Multiply this true divisor by the root figure last found, and subtract the product from the dividend. Seventh. Bring down the next period to the right of the remainder, to form the next dividend. Eighth. Double the part of the root already found for a trial divisor, and proceed is indicated in the 5th, 6th, 7th, and 8th of these paragraphs. What is the square root of each of the following num 217. Square Root of Fractions. (a.) Since the square of a fraction equals the square of its numerator divided by the square of its denominator, the square root of a fraction must equal the square root of its numerator divided by the square root of its denominator. (b.) If the numerator and denominator of a fraction are not perfect squares, we can only get the approximate value of its square root. (c.) In such cases, if the denominator is not a perfect square, it will be well to multiply both terms by such number as will make it so. This number may be either the denominator, or the product of the prime numbers which are found as factors in the denominator, 1, 3, 5, or any odd number of times. This differs from the true root by less than 2. (e.) Should a greater degree of accuracy be required, both terms may be multiplied by such a square number as will make the denominator sufficiently large to secure the requisite degree of accuracy. Thus, multiplying the numerator and denominator of 1659 22 X 32 X 73 What is the square root of each of the following frac 218. Square Root of Decimal Fractions. (a.) In order that a decimal fraction may have a perfect square for its denominator, it must have an even number of places in its numer ator. Thus, the denominators of .04, .25, .17, .6561, and .384736 are perfect squares, but the denominators of .4, 2.5, .017, .06561, .0384736, &c., are not. (b.) If a decimal fraction, the root of which is required, does not have an even number of decimal places in its numerator, a zero must be annexed, to make the number even, so that the denominator in all cases may be a perfect square. = .01, &c., it follows that there will be as many decimal places in the frac tional part of the root as there are times two decimal places in the fractional part of the power. Hence, we can carry out the root to as many places as we choose, by annexing two zeros to the power for every additional figure which we wish to obtain in the root. |
