CHAPTER III. Resolution of Problems and Equations of the Second Degree. 82. Introduction. When the enunciation of a problem leads to an equation of the form ax=b, in which the unknown quantity is multiplied by itself, the equation is said to be of the second degree, and the principles established in the two preceding chapters are not sufficient for the resolution of it; but b =- we a' since by dividing the two members by a, it becomes x2 see that the question is reduced to finding a number, which, b multiplied by itself, will produce the number expressed by This is the object of the extraction of the square root. The different procedures for extracting the square root of particular numbers, whether whole numbers or fractions, have been fully exposed in arithmetic; it is only necessary then to develope here the rules relative to the extraction of the roots of numbers, expressed algebraically. 1. Formation of the Square, and Extraction of the Square Root of Algebraic Quantities. 83. We will first consider the case of a monomial; and in order to discover the process, see how the square of the monomial is formed. By the rule for the multiplication of monomials, (16), we have 3 (5a3b3c)=5a2b3cx 5a2b3c=25a*b*c* ; that is, in order to square a monomial, it is necessary to square its coefficient, and double each of the exponents of the different letters. Hence, to find the root of the square of a monomial, it is necessary, 1st. To extract the square root of the coefficient by the rules given in arithmetic. 2d. To take the half of each of the exponents. Thus, √64ab48a3b2; for Sa3b2 × 8a3b2 =64a®b1. √625a2b3c=25ab1c3, for (25ab4c3)2=625a2bc®. From the preceding rule, it follows, in order that one monomial may be the square of another, its coefficient must be a perfect square, and all of its exponents even numbers. Thus, 98ab is not a perfect square, because 98 is not a perfect square, and a is affected with an uneven exponent. In this case, the quantity is introduced into the calculus by affecting it with the sign √ and it is written thus, √98ab1. Quantities of this kind are called radical quantities, or irrational quantities, or simply radicals of the second degree. 84. These expressions may sometimes be simplified, upon the principle that the square root of the product of two or more factors is equal to the product of the square roots of these factors; or, in algebraic language, Vabcd...= √a.√b.√c.√d........ To demonstrate this principle, we will observe, that from the definition of the square root, we have Again, (√abcd . ... .)2=abcd.... (√a× √b× √c× √d....)2=(√a)2 ×(√b)2 × (√c)2 ×( /d)3.. =abcd.... Hence, since the squares of Vabcd . . . ., and, √a. √b. √c. √d...., are equal, these quantities themselves are equal. This being the case, the above expression, √98ab*, can be put under the form √496x2a√496 x √2a. Now √4964 may be reduced to 762, (83); hence V98ab7b2. √2a. In like manner, √45a2b3c2d = √9a2b2 c2 × 5bd=3abc. √5bd, √864a2b3c11= √144ab4c10x6bc=12ab2cs. √6bc. In general, in order to simplify an irrational monomial, take all of those factors which are perfect squares, and extract the root of them, (83); then place the product of these roots before the radical sign, under which, leave those factors which are not perfect squares. In the expressions 7b2. √2a, 3abc. √ 5bd, 12ab2c3.√6bc, the quantities 7b2, 3abc, 12ab2c5, are called coefficients of the radical. 85. As yet, we have not paid any attention to the sign with which the monomial may be affected: but since in the resolution of questions we are led to consider monomial quantities preceded by the sign + or -, it is necessary to know how to operate upon quantities of this kind. Now the square of a monomial being the product of this monomial by itself, it follows (62) that, whatever may be its sign, the square of it will be positive. Thus, the square of +5ab3, or -5ab3, is +25a1b®. Whence we may conclude, that if a monomial is positive, its square root may be affected with either the sign + or; thus, √9a1 =±3a2, for +3a2 or -3a2, squared, gives 9a1. The double sign with which the root is affected is called plus or minu. If the proposed monomial were negative it would be impossible to extract its root, since it has just been shown that the square of every quantity whether positive or negative is essentially positive. Therefore, V9, √—4a2 √-5 are algebraic symbols which indicate operations that cannot be performed. They are called imaginary quantities, or rather imaginary expressions: they are symbols of absurdity, frequently met with in the resolution of equations of the second degree. These symbols can, however, by extending the rules, be simplified in the same manner as those irrational expressions which indicate operations that can be performed. Thus V-9 may be reduced to (84) √9x √-1 or 3√-1; √—4a2 = √ 4a2 × √—1=2a√—1; √-8a2b-4a2 x-2b=2a√-2b=2a√26x √-1. 86. Let us now try to discover the law of formation for the square of any polynomial whatever, from which a process may be deduced for extracting the root of this square. It has already been shown that the square of a binomial (a+b) is equal to a2 +2ab+b2. (19). Now to form the square of a trinomial a+b+c, denote a+b by the single letter s, and we have (a+b+c)2=(s+c)2 =s2 + 2sc + c2. But s2=(a+b)2=a2 +2ab+b2 ; and 2sc=2(a+b)c=2ac+2bc. Hence (a+b+c)2=a2 +2ab+b2+2ac+2bc+c2; that is, the square of a trinomial is composed of the sum of the squares of its three terms, and twice the product of these terms multiplied together two and two. This law of formation is true for any polynomial whatever. For suppose it verified for a polynomial of any number of terms, and see if it is true for one containing one more term. For this purpose, take the polynomial a+b+c+d...+i+k, containing m-1 terms, and denote the sum of the m first terms a+b+c+d.....+i by s; then s+k will represent the proposed polynomial, and we have (s+k)2 =s2 +2sk+k2; or substituting for s its value, (s+k)2=(a+b+c+d+ ... + i)2+2(a+b+c+d+... i)k+k3. But by hypothesis, the first part of this expression is composed of the squares of all the terms of the first polynomial and the double products of these terms taken two and two; the second part contains all of the double products of the terms of the first polynomial by the additional term k; and the third part is the square of this term. Therefore, the law of composition, announced above, is true for the new polynomial. But it has been proved to be true for a trinomial; hence it is true for a polynomial containing four terms; being true for four, it is necessarily true for five, and so on. Therefore it is general. This law can be enunciated in another manner: viz. The square of any polynomial contains the square of the first term, plus twice the product of the first by the second, plus the square of the second; plus twice the product of each of the two first terms by the third, plus the square of the third; plus twice the product of each of the three first terms by the fourth; plus the square of the fourth, and so This enunciation which is evidently comprehended in the on. first, shows more clearly the process for extracting the square root of a polynomial. From this law, (5a2 +4ab2)2=25a°—40a1b2 +16a2b1, or reducing, (3a2-2ab+4b3)2=9a1-12a3b+4a2b2+24a2b2-16ab3+16b1, =9a+12a3b+28a2b2 — 16ab3 +16b1, (5a2b-4abc+6bc2-3a2c)2=25a1b2-40a3b2c+76a2b2 c2 -48ab2c3+36b2c4-30abc+24a3bc2-36a2bc3+9a4c2. We will proceed to the extraction of the square root. Let the proposed polynomial be designated by N, and its root, which we will suppose is determined, by R; conceive, also, that these two polynomials are arranged with reference to one of the letters which they contain, a, for example. This being supposed, we will observe in the first place, that the two first terms of N (supposing it arranged) will give immediately the first and second terms of R; for it evidently follows, from the law of formation of the square, (86), 1st. That the exponent of the letter a, in the square of the first term of R, is greater than the exponent of this letter in any other of the parts which enter into the composition of the square of R. 2d. That the double product of the first term of R by the second, also contains a higher exponent than any of the following parts. Therefore the two parts just mentioned could not have been reduced with the others, and are necessarily the two terms of N, affected with the highest exponent of a, and the exponent immediately inferior to it; whence it follows, that if N is really a perfect square, 1st. Its first term must be a perfect square, and its root, extracted by the rule of No. 83, is the first term of R. 2d. Its second term must be divisible by twice the first term of R, and performing the division, the quotient will be the second term of R. To obtain the following terms, form the square of the binomial, already found, and subtract it from N; the remainder, which we will designate by N', contains the double product of the first term of R, by the third, plus some other parts. But the double product of the first term, by the third, must contain a, with a higher exponent than this letter has in any of the following |