term of the root, contained in the same term of the power, and divided by the number of terms to that place; the quotient will be the co-efficient belonging to the next term.* Note-To determine the signs of the different terms, it must be remembered that the even powers of negative quantities are positive, and the odd powers negative. Hence, a+x=a5+5a+x+10a3x2+10α2x2+5ax1+x3. Required the 6th power of 2x-3y. (2x)6 (2x)5.3y (2x)+.(3y)3 (2x)3.(3y)3 (2x)3.(3y)* 2x.(3y)3 (3y)6 Or, 64x6 96x3y 144x1y2 216x3y3 324x2y* 486xy3 729y6 Hence,(2x-3y)-64x6-576x5y+2160x1y2-4320x3y? +4860x3y*-2916xy3+729y®. 2. Required the 3d power of a+y. Ans. as+Say+зay3+y3. 3. Required the 4th power of a—b. Ans. a-4a3b+6a2b2-4ab3+b2. The co-efficients of any two terms, equally distant, the one from the beginning, and the other from the end, are alike; hence, the computation of the first half of them will supply the whole. 4. Required the 7th power of x+y. Result, x7+7xy+21x3y2+35x+y3 +, &c. 5. Required the 8th power of a-c. Result, as 8a7c+28aoc2—56a3c3+70a*c3, &c. 6. Required the 4th power of 2+x. Result, 16+32x+24x2+8x3+x2. 7. Required the Sd power of a-sb. Result, a3-9a2b+27 ab2-2763. This method of involution is easily extended to trinomials, quadrinomials, &c. by considering two or more of the terms as a single compound one. 8. Required the 4th power of a+b—c. Here considering b-c as the second term of the binomial, we have (a+b—c)*=a*+4a3(b—c)+6a2(b—c)= +4a(b-c)3+(bc); and involving b-c, to the powers indicated, by the same method, we have (b-c)-b3— 2bc+c; (b-c)3-b3-3boc+3bc2-c3; (bc)+ b2-4b3c +6b2c-4bc3+c; consequently, (a+b-c)=a*+4a3b -4a3c+6a2b2-12a2bc+6a2c2+4ab3-12ab2c+12abc2 4ac3+b+4b3c+6b2c2-4bc3+c+. 9. Required the Sd power of x+y+z. Result, x3-3x2y+3xy2+y3+3x2z+6xyz+ 3y2z+3xz2+3yz2+z3. 10. Required the 2d power of a+b-c―d. Result, a+2ab+b2-2ac-2bc+c2-2ad 2bd+2cd+do. 11. Required the 4th power of x+y-3z. Result, x+4x3y+6x2y2+4xу3+y1-12x3z36x2yz-56xyz-12y3z+54x2x2+108xyz2 +54y-108xz3-108yz8+812*. 12. Required the 5th power of a+b+c. Result, a5+5a+b+10a3b2+10a3b3+5ab1+b5 +5a+c+20a3bc+30a2bc+20ab3c+5b+c+ 10a3 c2+30a2bc2+30ab2c2+10b3c2+10a2c3 +20abc3+10b2c3+5ac2+5bc++c3. 36. To extract the root of a simple quantity. Extract the root of the co-efficient, for the numeral part; and divide the index or indices of the literal part, for the exponents of the root.* When the root cannot be extracted, it must be indicated, as in definition 16. EXAMPLES. 1. Required the 4th root of 256α*x. 4 8 4/256=4, the co-efficient; ax=ax2, the literal part. Hence, the root required is 4ax2, or —4ax2. 2. What is the 5th root of -5a15b10? The 5th root of 5 is a surd, and must be indicated thus: 5/5 or 1 53 Hence, the root required is —53a3b2 or —a3b35/5. * The odd roots have the same signs as their powers; but the even roots of positive quantities may be either positive or negative. The even roots of negative quantities are impossible. These impossible or imaginary roots, frequently become the subject of important investigations, as will appear in the sequel of this work. 3. Required the square root of 16a2b1. Result, +4ab3. 4. What is the square root of 625x+y3 ? Ans. ±3ab3, or ±ŝab✅b. 9. Required the 5th root of -243a5b. CASE 2. Result, -3ab5/b. 37. To extract the square root of a compound quantity. Arrange the terms according to the dimensions of some letter, beginning with the highest. Take the square root of the first term for the first term of the root, and subtract its square from the given quantity. Double the root thus found, for a defective divisor, divide the first term of the above remainder by this defective divisor, and annex the result both to the root and to the divisor. Multiply the divisor thus completed by the term of the root last obtained, and subtract the product from the former remainder. Divide the first term of the remainder as before, for the next term of the root; add this last and the preceding term of the root, to the last complete divisor for a new divisor. Multiply, subtract, and proceed as before. Note-In these additions, regard must always be paid to the signs of the quantities. EXAMPLES. 4a2+12ab+9620ac30bc+25c2(2a+3b-5c 2. Required the square root of x2+2xy+y3. Result, x+y. Result, a2+2a+1. 3. Required the square root of a1+4a3+6a2+4a+1. 4. Required the square root of x2-4ax3+6a2x2-4a3x +at. Result, a2-2ax+x2. 5. What is the square root of 16x2+24x3+89x2+60x +100 P Ans. 4x+3x+10. 6. What is the square root of 4a-12a3x+5a2x2+ 6ax2+x4? Ans. 2a2-3αx-x2. 7. What is the 4th root of a⭑+12a3b+54a2b2+108ab3 +8164 P* Ans. a+3b. 8. What is the 4th root of a3—4a®b2+6a+b2-4a2b¤ +68? CASE 3. Ans. a2b2. 38. To extract any root of a compound quantity. Arrange the terms as before directed; take the root of the first term for the first term of the root, and subtract its power from the given quantity. Take for a divisor, twice this root, three times its square, four times its third power, five times its fourth power, &c., according as the 2d, 3d, 4th, 5th, &c. roots are required to be extracted; divide the first term of the remainder, and annex the result to the root before found. * The 4th, 8th, 16th, &c. roots, may be obtained by 2, 3, 4, &c. extractions of the square root. |