SQUARE ROOT OF POLYNOMIALS. 230. To deduce a rule for the extraction of the square root of a polynomial, let us first observe how the square of any binomial, as a+b, is formed. We have (a+b)2 = a2+2ab+b2. And the last two terms may be written as follows: (2a+b)b. Let us now consider how the process of involution may versed, and the root, a+b, derived from the square. 2a+b be re OPERATION. a2+2ab+b2 a+b a3 2ab+b2 2ab+b Extracting the square root of a2, we obtain a, the first term of the root. Taking a from the whole expression, we have 2ab+b2, or (2a+b)b, for a remainder. Dividing the first term of this remainder by 2a, as a partial divisor, we obtain b, which we place in the root, and also at the right of the 2a to complete the divisor, 2a+b. Multiplying the complete divisor by b, and subtracting the product from the dividend, we have no remainder, and the work is finished. By the same process continued, we may extract the root of any quantity that is a perfect square. To establish the rule in a general manner, let a+b+c+d.... represent any polynomial. By a previous article, the square of this polynomial consists of the square of each term, together with twice the product of each term by all the terms which follow it ; (217); and the square may be written as follows: ... a2+2ab+2ac+2ad..... +b2+2bc+2bd.... +c2+2cd.. +d3.... And it is evident that if the root, a+b+c+d...., is arranged according to the powers of some letter, the square will also be arranged according to powers of the same letter. We may now derive the root from the square, in the following manner; OPERATION. a2+2ab+2ac+2ad. a2 2a+b 2ab+2ac+2ad....+b2+2bc+2bd....+c2+2cd....+ď2.......... We find a as in the former example, and take its square from the whole expression. We then divide the first term of the remainder by 2a, and write the quotient, b, in the root, and also in the divisor. We then multiply the complete divisor by b, subtract the product from the first remainder, and thus obtain a new dividend. Then writing 2a+26 for a partial divisor, we find c in the same manner as we found b; and thus we continue till the work is finished. If we examine the several subtrahends, taking the terms diag onally in the operation, we shall find a3, 2ab, 2ac, 2ad, etc.; b3, 2bc, 2bd, etc.; c2, 2cd, etc.; d3, etc. That is, we have, in the operation, the square of each term of the root, together with twice the product of each term by all the terms which follow it. Thus we have exactly reversed the process of forming a polynomial square. Hence the following general RULE. I. Arrange the terms according to the powers of some letter, and write the square root of the first term in the quotient. II. Subtract the square of the root thus found from the given quantity, and bring down two or more terms for a dividend. III. Divide the first term of the dividend by twice the root already found, and write the result both in the root and in the divisor. IV. Multiply the divisor, thus completed, by the term of the root last found, subtract the product from the dividend, and proceed with the remainder, if any, as before. NOTE-According to the law of signs in evolution, every square root obtained will still be a root, if the signs of all its terms be changed. 1? u? EXAMPLES FOR PRACTICE. 1. What is the square root of a2+2ab+2ac+b2+2bc+c2? Ans. a+b+c. 2. What is the square root of a*—6a2b+4a2 +96a −12b+4 ? Ans. a-3b+2. 3. What is the square root of x+4x+2x-2x2+5x2-2x+ Ans. x+2x2-x+1. 4. What is the square root of 1-2a+3a2-4a3+3a*—2a'+ Ans. 1-a+a—a3. 5. What is the square root of 4a*b*—12a3b2+8a3l3+9a3l3— 12a2b3+4a2b*? Ans. 2a2b-3ab+2ab3. 6. What is the square root of 9x°—30x*y+x*y2+76x3y3— 14x3y*—48xy*+36y° ? Ans. 3x-5x'y-4xy'+6y'. 7. What is the square root of a'-6a2bc+4a3cd-2a'd' +96*c* -12bc'd+6bcd'+4cd-4cd+d* ? Ans. a-3bc+2cd-d'. 9. What is the square root of x—6x*+11x3—6x ̄1+x−1? Ans. x-3x+x ̄3. 10. What is the square root of a2b¬2—10ab¬1+27—10a¬1b+ a-272? Ans. ab-5+a-1b. 2m 11. What is the square root of a1m+6a3¿"+11a2mc2n+6amc3n+ cin? Ans. am+3ac"+c3. 2m SQUARE ROOT OF NUMBERS. 231. In order to discover the process of extracting the square root of a number, it is necessary to determine 1st. The relative number of places in a number and its square root. 2d. The local relations of the several figures of the root to the periods of the number. 3d. The law by which the parts of a number are combined in the formation of its square. 232. The relative number of places in a given number and its square root may be shown by illustrations, as follows: From these examples we perceive that a root consisting of 1 place may have 1 or 2 places in the square; and that in all cases the addition of 1 place to the root adds 2 places to the square. Hence, If we point off a number into two-figure periods, commencing at the right hand, the number of periods will indicate the number of places in the square root. 233. If any number, as 2345, be decomposed at pleasure, the squares of the parts, beginning with the highest order, will be related in local value as follows: 20002 = 4 00 00 00 23002 = 5 29 00 00 23402 = 5 47 56 00 2345' = 5 49 90 25. Hence The square of the first figure of the root is contained wholly in the first period of the power; the square of the first two figures of the root is contained wholly in the first two periods of the power; and so on. 234. If the figures of a number be separated into two parts, and written with their local value, we may then form the square of the number by the formula for a binomial square. Thus, 7670 + 6. And if we put a = 70 and b = 6, then Hence, the binomial square may be used as a formula for extracting the square root of a number. 1. Let it be required to extract the square root of 5776. There are two periods in the number, indicating that there will be two places in the root. As the square of the tens is contained wholly in the first period, (233), we first seek the greatest perfect square in 57. This = OPERATION. 57 76 76 a', 49 00 2a, 140 876 2a+b, 146 876 we find to be 49, the root of which is 7, the first figure of the root sought. Hence we have a 70, and subtracting a2, or 4900, from the entire number, we have 876 for a remainder, which must be equal to (2a+b)b; (230). Dividing the remainder by the partial divisor, 2a, or 140, we have b = 6, the second figure of the root. Completing the divisor, we have 2a+b = 146 ; whence (2a+b)× b = 876, and the work is complete. It is obvious that we may omit ciphers, and still employ the figures with their proper local values, in the operation. It will not then be necessary to form the partial divisor separate from the complete divisor. If the given number consists of more than two periods, we may extract the two superior figures of the root from the first two periods, (233), bringing down another period to the remainder. Then a in the binomial formula will represent the part of the root already found, considered as tens of the next inferior order; and Having found 47, the square root of the first two periods, we bring down the last period, and have 5676 for a new dividend. We |