20 What is the second root of (2 m 21 What is the 6th root of (3 a + x)m ? x)? XXXVII. Roots of Compound Quantities. When a compound quantity is a perfect power, its root may be found; and when it is not a perfect power, its root may be found by approximation, by a method similar to that employed for finding the roots of numeral quantities. First we may observe, that no quantity consisting of only two terms can be a complete power; for the secor.d power of a binomial consists of three terms; that of a + x, for example, is a2 + 2 a x + x2. The quantity a2 + b2 is not a complete second power. Let it be required to find the second root of 9 x1 aε + 4 a2 ba + 12 x2 a1 b2. The se This The root of this will consist of at least two terms. cond power of the binomial a + b is a2+2ab+b2. shows that the quantity must be arranged according to the powers of some letter as in division, for the second power of either term of the root will produce the highest power of the letters in that term. Arrange the above according to the powers of x. 9 x* a° + 12 x2 a1 b2 + 4 a2 b1 ̈ ́ The formula a +2 a b + b2 shows that we should find the first term a of the root by taking the root of the first term; the same must be the case in the given example. The root of 9 x1 a is 3 x2 a3. Write this in the place of a quotient, and subtract its second power. Then multiply 3 x2 α3 by 2 for a divisor, answering to 2 a of the formula. 9 xˆ ao -† 12 xa a*b2 +4 a3 b1 (3° x2 a3 + 2 a b3 Divide the next term by the divisor. This gives 2 a b2 for the next term of the root. Raise the whole root then to the second power and subtract it. Or, which is the same thing, since the second power of the first term has already been subtracted, write the quantity 2 ab at the right of the divisor as well as in the root. Multiply the whole divisor as it then stands by the last term of the root. This produces the terms corresponding to 2 ab + b2, = b (2 a + b) of the formula. This produces 12x2 a b2+4 ab, which being subtracted, there is no remainder. Consequently the root is 3x2 a3 + 2 a b2 or 3 x2 a3-2 a b2. The second power of both is the same. the double sign had been given to the first term of the root, the second would have had it also, and the positive and negative roots would have been obtained together. Let it be required to find the 2d root of 36 a3 m*-60 a b m2 + 25 b2. 36 a2 m2 60 a b m2 + 25 b2 (6 am2 - 5 b ― If The process in this case is the same as in the last example. The second term of the root has the sign in consequence of the term 60 a b m2 of the dividend being affected with that sign. If the quantity had been arranged according to the powers of the letter b, thus, 25 b2 · 60 a b m2 + 36 a m1, the root would have been 5 b. 6.a m2 instead of 6 a m2 5 b. Both roots are right, for the second powers of the two quantities are the same. α. The second power a→→ b is the same as that of b One is the positive and the other the negative root. If the double sign be given to the first term of the root, both results will be produced at the same time in either arrangement. 25 b2- 60 a b m2 + 36 a2 m2 (±5b6am2 2562 * -60 a b m2 +36 a2 m2 (±10b 6am2 the + to—, When the quantity whose second root is to be found, consists of more than three terms, it is not the second power of a binomial, but of a quantity consisting of more than two terms Suppose the root to consist of the three terms m+n+p. If we represent the two first terms m+n by l, the expression becomes +p, the second power of which is l2 + 21p+p2. Developing the second power 12 of the binomial m+n, it becomes m2+2 m n + n2. This shows that when the quantity is arranged according to the powers of some letter, the second root of the first term will be the first term m of the root. If m2 be subtracted, and the next term be divided by 2 m, the next term n of the root will be obtained. If the second power of m + n or 12 be subtracted, the remainder will be 2 lpp. If the next term 21p be divided by 2 equal to twice m +n, the quotient will be p, the third term of the root. The same principle will extend to any number of terms. It is required to find the second root of 4 a* + 12 a3 x + 13 a2 x2 + 6 α x3 + x2. Let this be disposed according to the powers of a or of z. 3 x2 + 6 û x3- 13 a2x2 + 12 a3 x + 4 α* (x2 + 3 a x + 2 a2 root. 204 * 1st dividend. 6 a x3 + 13 a2 x2 (2x2+3ax 1st divisor. 6 ax3 9 a2 x2 2d divid. * 4a2x2 + 12 a3 x + 4 a1 (2x2 + 6 a x + 2 a 2d. di The process is so similar to that of numeral quantities that it needs no farther explanation. The double sign need not be given to the terms during the operation. All the signs may be changed when the work is done, if the other root is wanted. This will seldom be the case when all the terms are positive; but when some of the terms are negative, if it is not known which quantities are the largest, the negative root is as likely to be found first as the positive. When this happens the positive will be found by changing all the signs. Examples. 1. What is the second root of 4a3 x + 6 a2x2 + a* + x2 + 4 a x3? 3. What is the second root of -4x+4x+ 12 x3 — 6 x + x2 + 9? 4. What is the second root of x+20x25x+16+4x+10x+24 x? XXXVIII. Extraction of the Roots of Compound Quantities of any Degree. By examining the several powers of a binomial, and observing that the principle may be extended to roots consisting of more than two terms, we may derive a general rule for extracting roots of any degree whatever. (a + x)' = a + x a + x a2 + a x ax + x2 (a + x)2 = a2 + 2 α x + x2 ɑ + x a3 + 2 a2 x + a x2 a2 r + 2 x x2 + 203 (a + x)3 = a3 +3 a2 x + 3 a x2 + x3 а + x a'+3 a3x +3 a2x2 + a x3 a3x+3a2x2 + 3 α x3 + x* (a + x)* = a* + 4 a3 x + 6 a3 x2 + 4 α x2 + x2 a + x a2 + 4 a*x + 6 a3 x2 + 4 a2x2 + a x1 a*x + 4 a2x2 + 6 a2x2 + 4 α x2 + x0 (a + x)3 = ao + 5 a* x + 10 a3 x2 + 10 a2 x2 + 5 a xa +xa By examining these powers, we find that the first term is the first term of the binomial, raised to the power to which the binomial is raised. The second term consists of the first term of the binomial one degree lower than in the first term, multi |