To extract the square root of a compound Quantity. RULE. 238. Observe in what manner the terms of the root may be derived from those of the power; and arrange the terms accordingly; then set the root of the first term in the quotient; subtract the square of the root, thus found, from the first term, and bring down the next two terms to the remainder for a dividend. Divide the dividend, thus found, by double that part of the root already determined, and set down the result both in the quotient and divisor. Multiply the divisor, so increased, by the term of the root last placed in the quotient, and subtract the product from the dividend, and to the remainder bring down as many terms as are necessary for a dividend, and continue the operation as before. Ex. 1. Required the square root of a2+2ab+b2? a2+2ab+b2 On comparing a+b with a2+2ab+b2, we observe that the first term of the power (a2) is the square of the first term of the root (a). Put a therefore for the first term of the root, square it, and subtract that square from the first term of the power. Bring down the other two terms 2ab+b2, and double the first term (a) of the root; set down 2a, and having divided the first term of the remainder (2ab) by it, we have b, the other term of the root; and since 2ab+b2-(2a+b) xb, if to 2a the term b is added, and this sum multiplied by b, the result is 2ab+b2; which being subtracted from the terms brought down, nothing remains. Ex. 2. Required the square root of a2+2ab+b2+2ac+2bc +c2? a2+2ab+b2+2ac+2bc+c2(a+b+c a2 2a+b12ab+b2 2a+2b+c12ac+2bc+c2 2ac+2bc+c2 On comparing the root a+b+c, thus found with its power, the reason of the rule for deriving the root from the power is evident. And the method of operation is the same as in the last example. Thus, having found the first two terms of the root as before, we bring down the remaining three terms 2a +2bc+c2 of the power, and dividing 2ac by 2a, it gives c, the third term of the root. Next, let the last term (b) of the preceding divisor be doubled, and add c to the divisor thus increas ed, and it becomes 2a+2b+c; multiply this new divisor by c, and it gives 2ac+2bc+c2, which being subtracted from the terms last brought down, leaves no remainder. In like manner the following Examples are solved. 89 Ex. 3. Required the square root of 4x+6x3+ —x2+15x +25? 89 3 4x+6x+ x2+15x+25 (2x2+ 2x+5 4 Ex. 4. Required the square root of x+4x+2x+9x2—4x +4. Ans. x+2x2-x+2. Ex. 5. Required the square root of x2+4ax3+6a2x2+4a3x +a. Ex. 7. Required the square root of 4a+12a3x+13a2x2+ 6ax3+x1. Ans. 2a+3ax+x3. Ex. 8. Required the square root of 9x+12x+3+x2+20x +25. Ans. 3x2+2x+5. Ex. 9. Required the square root of a2+2ab+b2+2ac+ 2bc+c+2ad+2bd+2cd+d2. Ans. a+b+c+d. Ex. 10. Required the square root of a+12ab+54a2b2+ 108ab3+81b1. Ans. a2+6ab+9b2. Ex. 11. Required the square root of a-6a3x+15a2x220a3x3+15a2xa1—6ax3+xo. Ans. a3-3a2x+3ax2—x3. Ex. 12. Required the square root of a1—2a2x2+xa. Ans. a2-2. CASE III. To extract the cube root of a compound Quantity. RULE. 239. Arrange the terms as in the last case; and set the root of the first terms in the quotient; subtract the cube of the root, thus found, from the first term, and bring down three terms for a dividend. Next, divide the first term of the dividend by 3 times the square of that part of the root already determined, and set the result in the quotient; then, to 3 times the square of that part of the root, annex 3 times the product of the same part and the last result, and also the square of the last result, with their proper signs; and it will give the divisor, multiply the divisor by the term of the root last placed in the quotient, and subtract the product from the dividend, bring down three terms or as many as may be necessary for a dividend, and proceed as before. Ex. 1. Required the cube root of a3+3a2b+3ab2+b3 ? a3+3a2b+3ab2+b3 The reason of the rule may be made evident from a comparison of the roots with its cube. Or, thus, if the quantity whose root is to be extracted, has an exact root, the root of the leading term must be one term of its root; that is, the cube root of a3, which is a, is one term of the root, and the remaining terms being brought down, the root of the last term 63 is consequently another term of the root; but as the root may consist of more terms than two; the next term (b) of the root is always found by dividing 3a2b 3a2 =b) the first term of the dividend by three times the square of the divisor, and the two remaining terms of the dividend 3ab2+b=(3ab+b2)b; hence 3ab+b must be added to 3a2 for a divisor; and so on. Ex. 2. Required the cube root of x+6x5-40x3-96x-64. 2+625-40x3+96x-64 (x2+2x-4 3x+6x+4x2)6x-40x3 6x3+12x1+8x3 3x+12x3-24x+16)-12x2-48x3+96x-64 -12x-48x3+96x-64 Ans. a+b+c. Ex. 3. Required the cube root of (a+b)3 + 3(a+b)2c+ 3(a+b)c2+c3. Ex. 4. Required the cube root of x-6x+15-20+ 15x2-6x+1. Ex. 5. Required the cube root of x+6x5y+15x'y2+20x1y3 +15x3y1+6xy3+yo. Ans. x2+2xy+y2. Ans. x2-2x+1. Ex. 6. Required the cube root of 1-6x+12-8x3. CASE IV. Ans. 1-2x. To find any root of a compound Quantity. RULE. 240. Find the root of the first term, which place in the quotient; and having subtracted its corresponding power from that term, bring down the second term for a dividend. Divide this by twice the part of the root above determined, for the square root; by three times the square of it, for the cube root; by four times the cube of it, for the fourth root, &c. and the quotient will be the next term of the root. Involve the whole of the root, thus found, to its proper power, which subtract from the given quantity, and divide the first term of the remainder by the same divisor as before. Proceed in the same manner, for the next following term of the root; and so on, till the whole is finished. 241. This rule may be demonstrated thus; (a+b)"=a® +nan-1b+ &c. Here the nth root of an is a, and the next term nan-1b contains b, (the other term of the root) nan-1. times; hence, if we divide nan-1b by nan-1, we have b, or nan--1b nan-i b; and so on, for any compound quantity, the root of which consists of more than two terms. Now, if n=2; then, the divisor nan-1=2a, for the square And so on for any other root, that is, involve the first term of the root, to the next lowest power, and multiply it by the index of the given power for a divisor. Ex. 1. Required the square root of a1-2a3x+3a2x2-2ax3 +24? a1-2a3x+3a2x2-2ax3-x1(a2—ax+x2 a1 2a2)-2a3x (a3—ax)2= àa—2a3x+a2x2 2a2)+2a2x2 (a2—ax+x2)2=a1 — 2a3x+3a2x2—2ax3+xa Ex. 2. Required the 4th root of 16a-96a3x+216a2x2— 216ax3+81x4. 16a-96a3x+216a2x2-216ax3+81x1(2a-3x 16a 4X(2a)=32a3)-96a3x (2a-3x)=16a-96a3x+216a2x2-216ax2+81x* 242. As this rule, in high powers, is often found to be very laborious, it may be proper to observe, that the roots of certain compound quantities may sometimes be easily discovered: |