2. Find the root of a* + 4a3b + 10a2b2 + 12ab3 + ba. a2 + 4a3b + 10a2b2 + 12ab3 + ba ( a2 + 2ab + 3b". 3. To find the square root of a + 4a3 + 6a2 + 4a + 1. To find the Roots of any Powers in General. THIS is also done like the same roots in numbers, thus: Find the root of the first term, and set it in the quotient. -Subtract its power from that term, and bring down the second term for a dividend.-Involve the root, last found, to the next lower power, and multiply it by the index of the given power, for a divisor.-Divide the dividend by the divisor, and set the quotient as the next term of the root.Involve now the whole root to the power to be extracted; then subtract the power thus arising from the given power, and divide the first term of the remainder by the divisor first found; and so on till the whole is finished *. EXAMPLES. * As this method, in high powers, may be thought too laborious, it will not be improper to observe, that the roots of compound quantities may sometimes be easily discovered, thus: Extract the roots of some of the most simple terms, and connect them together by the sign + or as may be judged most suitable for the purpose.-Involve the componnd root, thus found, to the proper power; then, if this 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. to +, Thus, EXAMPLES. 1. To find the square root of a1-2a3b+3a2b2 — 2ab3 + b*. a2-2a3b+3a2b2 - 2ab3 + b2 (a2 — ab + b2 a4 2a2) -2a3b a^~2a3b + a2b2 = (a2-ab) 2a2) 2a2b2 aa—2a3b + 3a2b2 —2ab3 + ba = (a2 — ab + b2)2. 2. Find the cube root of ao- 6a5 + 21aa — 44a3 + 63a2-54a + 27. a-6a3 +21a-44a3 + 63a2-54a+27 ( a2-2a + 3. a-6x+21a4-44a3+63a2 - 54a +27 (a2-2a-3)3. 2ab+2ax + b2 3a59a413a3 + 18a2 Ans. aa+2. 5. Find the 4th root of 81a4 216a3b+216a2b2 - 96ab3 + 1664. Ans. 3a-2b. 6. Find the 5th root of a5 - 10aa + 40a3 — 80a2 + 80n · -32. 7. Required the square root of 1 − x2. 8. Required the cube root of 1-x3. . Ans. a-2. Thus, in the 5th example, the root 3a-26, is the difference of the roots of the first and last terms; and in the 3d example, the root a-b+x, is the sum of the roots of the 1st, 4th, and 6th terms. The same may also be observed of the 6th example, where the root is found from the first and last terms. SURDS. I SURDS are such quantities as have no exact root; and are usually expressed by fractional indices, or by means of the radical sign. Thus, 3, or 3, denotes the square root of 3; and 23 or 3/22, or 3/4, the cube root of the square of 2; where the numerator shows the power to which the quantity is to be raised, and the denominator its root. PROBLEM I. To Reduce a Rational Quantity to the Form of a Surd. RAISE the given quartity to the power denoted by the index of the surd; then over or before this new quantity set the radical sign, and it will be of the form required. EXAMPLES. 1. To reduce 4 to the form of the square root. I then 3274° or (27a) is the answer. 3. Reduce 6 to the form of the cube root. 4. Reduce ab to the form of the Ans. (216) or 216. Ans.22. square root. 5. Reduce 2 to the form of the 4th root. Ans. (16). 6. Reduce a3 to the form of the 5th root. PROBLEM II. To Reduce Quantities to a Common Index. 1. REDUCE the indices of the given quantities to a common denominator, and involve each of them to the power denoted by its numerator; then I set over the common denominator will form the common index. Or, 2. If the common index be given, divide the indices of the quantities by the given index, and the quotients will be the new indices for those quantities. Then over the said quantities, with their new indices, set the given index, and they will make the equivalent quantities sought I EXAMPLES. 1. Reduce 3 and 53 to a common index. Here and and . 57%=(35)Tʊ Therefore 3% and 5% = (35)ro and (52) = 1935 and 1952 = 1/243 and 25. 2. Reduce a3 and 63 to the same common index . Here,+= × = the 1st index, and += x = the 2d index. Therefore (a) and (63)3, or √/a° and tities. b3 are the quan 3. Reduce 43 and 5 to the common index 4. I Ans. 2563) and 254. 4. Reduce a3 and to the common index . I Ans. (a2) and (.xa3)%. 5. Reduce a2 and r3 to the same radical sign. Ans. a and √x. 6. Reduce (a'+ x)3 and (a—x) to a common index. 7. Reduce (a + b) and (a—b)3 to a common index. PROBLEM III. To Reduce Surds to more Simple Terms. FIND out the greatest power contained in, or to divide the given surd; take its root, and set it before the quotient or the remaining quantities, with the proper radical sign between them. 1. To reduce Here √32 = EXAMPLES. 32 to simpler terms. 16 x 2 = √16 × √2 = 4 × √2 = 4√2. 2. To reduce 3/320 to simpler terms. 3/320 = 3/64 x 5 = 3/64 × 3/5 = 4 × 3/5 = 43/5. 3. Reduc 3. Reduce/75 to its simplest terms. Note. There are other cases of reducing algebraic surds to simpler forms, that are practised on several occasions; one instance of which, on account of its simplicity and usefulness, may be here noticed, viz. in fractional forms having compound surds in the denominator, multiply both numerator and denominator by the same terms of the denominator, but + having one sign changed, from to or from to +, which will reduce the fraction to a rational denominator. √20+ √12 -- √/5+ √ 3 Ex. To reduce √5-√3 multiply it by and 16+2/15 √5+√3' it becomes =8+15. Also, if 2 √/15-√5 √15 +5 657/75 multiply it by and it becomes 15-5' 15-5 65-35/3 13-7/3 10 PROBLEM IV. To add Surd Quantities together. 1. BRING all fractions to a common denominator, and reduce the quantities to their simplest terms, as in the last problem.-2. Reduce also such quantities as have unlike indices to other equivalent ones, having a common index.— 3. Then, if the surd part be the same in them all, annex it to the sum of the rational parts, with the sign of multiplication, and it will give the total sum required. But if the surd part be not the same in all the quantities, they can only be added by the signs + and -. 1. Required to add First, 189 x 2 EXAMPLES. √18 and √32 together. 3/2; and 32/16x2=4/2: Then, 3/2 + 4√/2 = (3 + 4) √27√2=sum required. 2. It is required to add 3/375, and 3/192 together. First, 3/375 3/125x3=53/3; and 3/1923/64 × 3=43/3: Then, 53/3+43/3 = (5+ 4) 3/3 = 93/3 = sum required. 3. Required |