performed. They are called imaginary quantities, or rather, imaginary expressions, and are frequently met with in the resolution of equations of the second degree. 127. Let us now examine the law of formation of the square of a polynomial; for, from this law, the rule is deduced for extracting the square root. It has already been shown (Art. 46), that, (a + b)2 = a2 + 2ab + b2; that is, The square of a binomial is equal to the square of the first term plus twice the product of the first term by the second, plus the square of the second. The square of a polynomial, is the product arising from multiplying the polynomial by itself once hence, the first term of the product, arranged with reference to a particular letter, is the square of the first term of the polynomial arranged with reference to the same letter. Therefore, the square root of the first term of such a product will be the first term of the required root. 128. Let us now extract the square root of the polynomial 28a5 + 49a4 + 4a6 + 9 + 42a2 + 12a3, which arranged with reference to the letter a, becomes, 4a5 + 28a5 + 49a2+12a3+42a2+92a3+ 7a2 +3 4a6 4a3+ 7a2 7a2 28a5+49a4 = (2r + r) r 12a3 + 42a2 +94a3+14a2+3 0 3 012a2+42a2+ 9 = (2n + r') Now, since the square root of 4a6 is 2a3, it follows that 2a3 is the first term of the required root. Designate this term by r, and the following terms of the root, arranged with reference to a, by r', r', r", &c. Now, if we denote the given polynomial by N, we shall have N = (r + r' + r'" + r'" + &c. ;)2 or, if we designate all the terms of the root, after the first, by s 2 N = (r + s)2 = r2 + 2rs + s2 = r2 + 2r (r' + r' + r'" + &c.) + s2. If now we subtract r2 = 4a6, from N, and designate the remainder by R, we shall have R = N - 4a = 2r (r + r' + r''" + &c.) + s2; in which the first term 2rr will contain a to a higher power than either of the following terms. Hence, if the first term of the first remainder be divided by twice the first term of the root, the quotient will be the second term of the root. and designate the remaining terms of the root, r", r'", &c., by s, we shall have R' = N - n2 = (2r + 2r') (r'" + r'" + &c.) + 8/12 ; in which, if we perform the multiplications indicated in the second member, the term 2rr" will contain a higher power of a than either of the following terms. Hence, if the first term of the second remainder be divided by twice the first term of the root, the quotient will be the third term of the root. If we make r + r' + r' = n', and r'"' + rIV + &c. = s", we shall have N = (n' + s")2 = n'2 + 2n's''+ s2; and R" = N - n'2 = 2 (r + r + r') (r'" + rv + &c.) + s2; from which we see, that the first term of any remainder, divided by twice the first term of the root, will give a new term of the required root. It should be observed, that instead of subtracting n2 from the given polynomial, in order to find the second remainder, that that remainder could be found by subtracting (2r + r')r' from the first remainder. So the third remainder may be found by subtracting (2n + r'") r" from the second, and similarly for the remainders which follow. In the example above, the third remainder is equal to zero, and hence the given polynomial has an exact root. Hence, for the extraction of the square root of a polynomial, we have the following ८ I. Arrange the polynomial with reference to one of its letters, and then extract the square root of the first term, which will give the first term of the root. Subtract the square of this term from the given polynomial. II. Divide the first term of the remainder by twice the first term of the root, and the quotient will be the second term of the root. III. From the first remainder subtract the product of twice the first term of the root plus the second term, by the second term. IV. Divide the first term of the second remainder by twice the first term of the root, and the quotient will be the third term of the root. V. From the second remainder subtract the product of twice the first and second terms of the root, plus the third term by the third term, and the result will be the third remainder, from which the fourth term of the root may be found; and proceed in a similar manner for the remaining terms of the root. EXAMPLES. Extract the square root of the polynomial 24ab3 + 25a 30a3b + 1664. 1664) (1. 49a2b2 First arrange it with reference to the letter a. (5. 5. Find the square root of 9a4 12a3b + 28a2b2 - 16ab3 + 6. Find the square root of 1664) 25a462 - 40a3b2c + 76a2b2c2 — 48ab2c3 + 3662c4 30abc + 24a3bc2 36a2bc3 + 9a4c2. 129. We will conclude this subject with the following remarks: 1st. A binomial can never be a perfect square. For, its root cannot be a monomial, since the square of a monomial will be a monomial; nor can the root be a polynomial, since the square of the simplest polynomial, viz., a binomial, will contain at least three terms. Thus, an expression of the form a2 + b2 can never be a perfect square. 2d. A trinomial, however, may be a perfect square. If so, when arranged, its two extreme terms must be squares, and the middle term double the product of the square roots of the other two. Therefore, to obtain the square root of a trinomial, when it is a perfect square, Extract the roots of the two extreme terms, and give these roots the same or contrary signs, according as the middle term is positive or negative. To verify it, see if the double product of the two roots is equal to the middle term of the trinomial. Thus, for, and also, But 9a6 - 48a4b2 + 64a2b4 is a perfect square, 9a6 = 3a3, and √64a2b4 = 8ab2, 2 × 3a3 x 8ab2 = 48a462, the middle term. 4a2+14ab + 962 is not a perfect square: for, although 4a2 and + 962 are the squares of 2a and 3b, yet 2 × 2a × 36 is not equal to 14ab. 3d. When, in extracting the square root of a polynomial, the first term of any one of the remainders is not exactly divisible by twice the first term of the root, we may conclude that the proposed polynomial is not a perfect square. This is an evident consequence of the course of reasoning, from which the general rule for extracting the square root was deduced. 4th. When the polynomial is not a perfect square, the expres sion for its square root may sometimes be simplified. Take, for example, the expression √a3b + 4a2b2 + 4ab3. The quantity under the radical is not a perfect square; but it can be put under the form ab (a2 + 4ab + 462). Now, the factor within the parenthesis is evidently the square of a + 2b, whence we have a3b + 4a2b2 + 4ab3 = (a + 2b) ab. Of the Calculus of Radicals of the Second Degree. 130. A radical quantity is the indicated root of an imperfect power. If the root indicated is the square root, the expression is called a radical of the second degree. Thus, √a, 36, 7√2, are radicals of the second degree. 131. Two radicals of the second degree are similar, when the quantities under the radical sign are the same in both. Thus, 3√b and 5c o are similar radicals; and so also, are 9√2 and 7√2. Addition and Subtraction. 132. In order to add or subtract similar radicals, add or subtract their co-efficients, and to the sum or difference annex the common radical. and 5cb = (3a – 5c) √b. 7√2a+3√2a = (7 + 3) √ 2a = 10 √2a; 3 = (7 3) √2a = Two radicals, which do not appear to be similar, may become so by simplification (Art. 125). |