Ex. 2. Required the square root of a +2ab+b+2ac+2bc +c ? a2+2ab+b2+2ac+2bc+c2(a+b+c a2 2a+b | 2ab+b2 2a+2b+c | 2ac+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 increased, 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+ 15.x +25? 3 4x2+6x2+ 2x2+15x+25 (2x2+ 2x+5 Ex. 4. Required the square root of x+4x+2x1+9x2-4x +4. Ans. x+2x2x+2. Ex. 5. Required the square root of x+4x+6a2x2+4a3x Aus. 2+2ax+a2. +a1. Ex. 6. Required the square root of a-2a3+3a2 —‡a+;}. Ans. a2-a+1. Ex. 7. Required the square root of 4a+12a3x+13a2x2+ 6ax2+x1. Ans. 2a2+3ax+x2. Ex. 8. Required the square root of 9x+12x+34x2+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+8164. Aus. a2+6ab+9b2. Ex. 11. Required the square root of ao—6a3x+15a3x2— 20a3x3+15a2x2—6ɑx3+x6. Ans. a3-3a2x+3ax2—x3. Ex. 12. Required the square root of a1-2a2x2+x1. CASE III. Ans. a2x2. To extract the cube root of a compound Quantity. RULE. 300. Arrange the terms as in the last case; and set the root of the first term 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. an Or, thus, if the quantity whose root is to be extracted, has 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 b3 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 the first term of the dividend by three times the 3a2 square of the divisor, and the two remaining terms of the dividend Sab+b3=(3ab+b2)b; hence 3ab+b2 must be added to Sa2 for a divisor; and so on. 6) Ex. 2. Required the cube root of x+6x-40x96x-64. x+6x-40x3+96x-64 (x2+2x-4 x6 3x+6x+4x2)6x5-40x3 6x+12x+8x3 3x+12x3-24x+16)-12x-48x3+96x-64 -12x2-48x3+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+15x-20x3+ 15x2-6x+1. Ex. 5. Required the cube root of x+6xy+15x1y2+20x3 y3+15x2y+6xy3+yo. Ans. x2+2xy+y2. Ans. x-2x+1. Ex. 6. Required the cube root of 1-6x+12x2-8x3. CASE IV. Ans. 1-2x. To find any root of a compound Quantity. RULE. 301. 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. 302. This rule may be demonstrated thus ; (a+b)"=a"+ nan-1b+&c. (Art. 289). 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 nan-1b have b, or b; and so on, for any compound quantinan-i tity, the root of which consists of more than two terms. Now, if n=2; then, the divisor na11=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 +x2? a1—2a3x+3a2x2-2ax2+x2(a2—ax+x2 2a3)-2a3x (a2-ax)2=a1-2a3x+a2x2 2a2)+2a2x2 (a2—ax+x2)2=aa—a3x+3a2x2-2ax3+xa Ex. 2. Required the 4th root of 16a-96a3x+216a2x2216ax+81x. 16a-96a3x+216a2x2-216ax3+81x+(2α-3x 16a1 4× (2a)3=32a3)—96a3x (2a—3x)1—16a2—96a3x+216a2x2-216ax3+81xa 303. 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 discover ed: thus, in the last example, the root is 2a-3x, which is the difference of the roots of the first and last terms; and so on, for other compound quantities. to Hence, the following method in such cases; extract the roots of all the simple terms, and connect them together by the signs or, as may be judged most suitable for the purpose then involve the compound root thus found, to its proper power, and if it be the same with the given quantity, it is the root required. But if it be found to differ only in some of the signs, change them from + to, or from +, till its power agrees with the given one throughout. However, such artifices are not to be used by learners, because the regular mode of proceeding is more advantageous to them; besides, a knowledge of those artifices which are used by experienced Algebraists, can only be acquired from frequent practice. Ex. 3. Required the square root of a2+2ab+b2+2ac+2bc +c2. Here, the square root of a2=a; the square root of b2=b and the square root of c2=c. Hence, a+b+c, is the root required, because (a+b+c)2=a2+2ab+b2+2ac+2bc=c2. Ex. 4. Required the fifth root of 32x-80x+80x3- 40x2 +10x-1. Ans. 2x-1. Ex. 5. Required the cube root of x-6x+15xa—201⁄23+ 15x2-5x+1. Ans. x2-2x+1. Ex. 6. Required the fourth root of a-4a3x+6a2x2-4ax3 +x1. Ex. 7. Required the square root of x3+2x1y*+y3. Ans. a-x. Ans. x+y. Ex. 8. Required the square root of x3 — 2x1y1+y3. Ans. x-y1. Ex. 9. Required the cube root of a3-6a2x+12ax2-8x3. Ans. α-2x. Ans.x-1. Ex. 10. Required the sixth root of xo—6x5+15xa—20x3+ 15x2-6x+1. Ex. 11. Required the fifth root of 210+1523y2+90x©y*+ 270x*y®+405x2y3+243y1o. Ans 22+3y2. Ex. 12. Required the square root of a2+2xy+y2+6xz+ Ans. x+y+3z. § III. INVESTIGATION OF THE RULES FOR THE EXTRACTION OF THE SQUARE AND CUBE ROOTS OF NUMBERS. 304. It has been observed, (Art. 104), that, a denoting the tens of a number, and b the units, the formula a2+2ab+b2 would represent the square of any number consisting of two |