= (x3-3x2)8-3(x3-3x2)2(2x+1)+3(x3-3x2) (2x+1)2 - (2x+1)* =x9-9x8+27x7-27x6-3(x-6x+9x+) (2x+1)+3(x3-3x2) (4x2+4x+1) -(8x8 +12x2+6x+1) =x9 9x8 +27 x7 - 27 x6 · 6 x + 33 x6 - 36 x5 - 27 x2 + 12 x5 — 24 x4 =x9 9x8+21 x 6 x6 — 24 x5 — 51 x - 41 x3- 21 x2 - 6 x − 1. CHAPTER XIII EVOLUTION 224. Evolution is the operation of finding a root of a quantity; it is the inverse of involution. 225. It follows from the law of signs in involution that: 1. Any even root of a positive quantity may be either positive or negative. 2. Every odd root of a quantity has the same sign as the quantity. √9 = +3, or −3 (usually written ±3); for (+3)2 and (−3)2 equal 9. 27 == 3, for (3)3 — — 27. 226. Since even powers can never be negative, it is evidently impossible to express an even root of a negative quantity by the usual system of numbers. Such roots are called imaginary numbers, and all other numbers are, for distinction, called real numbers. Thus 1 is an imaginary number, which can be simplified no further. EVOLUTION OF MONOMIALS 227. The following examples are solved by the definition of 228. To extract the root of a power, divide the exponent by the index. A root of a product equals the product of the roots of the factors. To extract a root of a fraction, extract the roots of the numerator and denominator. 31. √a2+2ab+b2. 35. V−32(m+n)3r. 39. √.49 p2. 229. A trinomial is a perfect square if one of its terms is equal to twice the product of the square roots of the two other terms. (§ 118.) In such a case the square root can be found by in spection. Ex. 1. Find the square root of x6 - 6 x3y2+9 y1. Hence x¤ — 6 x3y2 + 9 y1 = (x13 — 3 y2)2. (§ 118.) √x2-6x3y2+9y+=(x3-3y2). EXERCISE 89 Extract the square roots of the following expressions: 1. 1-4a+4 a2. 2. at +16 b2 - 8 a2b. 3. a+1-2 a2. 4. 16 at + ab + 8 a3. 5. 149 y 14 y2. 6. x2y2-6xyz +9 z2. 9. a+b8-2 a2b1. 10. 16 a1- 120 a2bc +225 b2c2. 13. a2+b2+ c2 + 2 ab + 2 bc + 2 ac. 230. In order to find a general method for extracting the square root of a polynomial, let us consider the relation of a + b to its square, a2 + 2 ab + b2. The first term a of the root is the square root of the first term a2. The second term of the root can be obtained by dividing the second term 2 ab by the double of a, the so-called trial divisor; a+b is the root if the given expression is a perfect square. In most cases, however, it is not known whether the given expression is a perfect square, and we have then to consider that 2 ab+b2=b(2 a+b), i.e. the sum of trial divisor 2 a, and b, multiplied by b must give the last two terms of the square. The work may be arranged as follows: a2+2ab+b2 | a+b a2 2a+b2ab+b2 2 ab+b2 Ex. 1. Extract the square root of 16 x2 - 24 x2y3 +9 y3. Explanation. Arrange the expression according to descending powers of x. The square root of 16 x is 4x2, the first term of the root. Subtracting the square of 4x2 from the trinomial gives the remainder - 24 x2y3 + 9 yo. By doubling 4x2, we obtain 8x2, the trial divisor. Dividing the first term of the remainder, 24 x2ys, by the trial divisor 8 x2, we obtain the next term of the root - 3y3, which has to be added to the trial divisor. Multiply the complete divisor 8 x2 3 y3 by - 3 y3, and subtract the product from the remainder. As there is no remainder, 4 x2 3 y3 is the required square root. |