— 5QR + 5T, Q, E=P5-5P3Q + 5P2R + 5PQ2 5PS &c. &c. which values may be continued on, at pleasure, by multiplying the last by P, the last but one by the last but two by R, the last but three by S, &c. and then adding all the products together; as is evident from the equations above derived. These conclusions are of use in finding the limits of equations, and contain a demonstration of a rule, given for that purpose, by Sir Isaac Newton, in his Universal Arithmetic. SECTION XII. Of the Resolution of Equations of several Dimensions. BEFORE we proceed to explain the methods of resolving cubic, biquadratic, and other higher equations, it will be requisite, in order to render that subject more' clear and intelligible, to premise something concerning the origin and composition of equations. Mr. Harriot has shown how equations are derived by the continued multiplication of binomial factors into each other according to which method, supposing x-a, x-b, x c, x-d, &c. to denote any number of such factors, the value of x is to be so taken, that some one of those factors may be equal to nothing: then, if they be multiplied continually together, their product must also be equal to nothing, that is, x— ах х -b × x xd &c. 0: in which equation, a may, it is plain, be = x equal to any one of the quantities, a, b, c, d, &c. since any one of these being substituted instead of x, the whole expression vanishes. Hence it appears, that an equation may have as many roots as it has dimensions, or as are expressed by the number of the factors, whereof it is supposed to be produced. Thus the quadratic equation b = 0 or x2 x + ab 0, has X -a x x a %} 11 two roots, a and b; the cubic equation a x abc= 0, has three roots, a, b, and c; and the biquadratic equation, xa xx-b has four roots, a, b, c, and d. From these equations it is observable, that the coefficient of the second term is always equal to the sum of all the roots, with contrary signs; that the coefficient of the third term is always equal to the sum of their rectangles, or of all the products that can possibly arise by combining them, two and two; that the coefficient of the fourth is equal to the sum of all their solids, or of all the products which can possibly arise, by combining them three and three; and that the last term of all, is produced by multiplying all the roots continually together. And all this, it is evident, must hold equally, when some of the roots are positive and the rest negative, due regard being had to the signs. Thus, in the cubic equation x -a x x · b × x + c = 0, or x3 + + ab ac x + abc = 0 (where two of the roots, a, b, are -bes positive, and the other, c, is negative, the coefficient of the second term appears to be ab+c, and that of the third, ab-ac-be, or ab + a × —c + b × -c, conformable to the preceding observations. Hence it follows, that, if one of the roots of an equation be given, the sum of all the rest will likewise be given; and that, in every equation where the second term is wanting, the sum of all the negative roots is exactly equal to that of all the positive ones; because, in this case, they mutually destroy each other. But when the coefficient of the second term is positive, then the negative roots, taken together, exceed the positive ones. But the negative roots, in any equation, may be changed to positive ones, and the positive to negative, by changing the signs of the second, fourth, and sixth terms, and so on alternately. Thus, the foregoing equation +ab a x x ·b × x + c =) x3 + —b } x2—ac}x+ +c) -bc 0, by changing the signs of the second and fourth +a +ab terms, becomes a3 + + bxc2. z ac abc x be = 0, or x + a × x + b x x c = 0; where the roots, from + a, b, and c, are now become a, —b, and + c. Moreover the negative roots may be changed to positive ones, or the positive to negative, by increasing or diminishing each, by some known quantity. Thus in the quadratic equation x2+8x + 150, where the two roots are 3 and 5 (and therefore both negative) if 7 be substituted for x, or which is the same, if each of the roots be increased by 7, the equation will become -712+8x=7+15= 0; that is, z2 — 6% +8 = 0, or z 2X 7 - 4 = 0; where the roots are 2 and 4, and therefore both positive. This method of augmenting, or diminishing the roots of an equation is sometimes of use in preparing it for a solution, by taking away its second term; which is always performed by adding, or subtracting 1, 3, or 4 part, &c. of the coefficient of the said term, according as the proposed equation rises to two, three, or four, &c. dimensions. Thus, in the quadratic equation x2- 8x + 15 = 0, let the roots be diminished by 4, that is, let x - 4 be put = ≈, or x = 4+; then, this value being substituted for x, the equation will become + 4128 x ≈ + 4 0, or z2 — 1 = 0; in which the second term is + 15 = wanting. Likewise, the cubic equation z3 — az2 + bz — c = 0, Hence it appears, how any affected quadratic may be reduced to a simple quadratic, and so resolved without completing the square; but this, by the bye. I now proceed to the matter proposed, viz. the resolution of cubic, biquadratic, and other higher equations; and shall begin with showing How to determine whether some, or all the roots of an equation be rational, and, if so, what they are. Find all the divisors of the last term, and let them be substituted, one by one, for x in the given equation; and then, if the positive and negative terms destroy each other, the divisor so substituted is manifestly a root of the equation; but if none of the divisors succeed, then the roots, for the general part, are either irrational or impossible: for the last term, as is shown above, being always a multiple of all the roots, those roots, when rational, must, necessarily, be in the number of its divisors. Examp. 1. Let the equation a34x2-7x+10= 0, be proposed; then, the divisors of (10) the last term being + 1, 1, +2, 2, +5, 5, +10, 10, let these quantities be, successively, substituted instead of x. and we shall have, 4— 7+10= 0, therefore 1 is a root; —1— 4+ 7+10=12, therefore -1 is no root; 8-16-14+10=-12, therefore 2 is no root; —8— 16+14+10= 0, therefore -2 is another root; 125-100-35+10= 0, therefore 5 is the third root. It sometimes happens that the divisors of the last term are very numerous; in which case, to avoid trouble, it will be convenient to transform the equation to another, wherein the divisors are fewer; and this is best effected by increasing or diminishing the roots by an unit, or some other known quantity. Examp. 2. Let the equation propounded be y1—4y3— 8y+32=0; and, in order to change it to another whose last term admits of fewer divisors, let x+1 be substituted \therein for y, and it will become x2-6x2-16x+21 0; where the divisors of the last term are, 1, -1, -7, 21, and -21; which being, successively, substituted for x, as before, we have, 3, 3, 7, 1- 6. 16+21=0, therefore 1 is one of the roots; 6 +16+21=32, therefore −1 is not a root; 54 48+21=0, therefore 3 is another root. But the other two roots, without proceeding further, will appear to be impossible; for, their sum being equal to -4, the sum of the two positive roots (already found,) with a contrary sign (as the second term of the equation is here wanting,) their product, therefore, cannot be equal to (7) the last term divided by the product of the other roots, as it would, if all the roots were possible. However, to get an expression for these imaginary roots, let either of them be denoted by v, and the other will be denoted by 4-v; which, multiplied together, give —4v -27; whence v-2+-3, and consequently -4 == -2-3. Now let each of the four roots found above, be increased by unity, and you will have all the roots of the equation proposed. When the equation given is a literal one, you may still proceed in the same manner, neglecting the known quantity and its powers, till you find what divisors succeed; for each of these, multiplied by the said quantity, will be a root of the equation. Thus, in the literal equation x3+3ax2-4a2x 12a3 0, the numeral divisors of the last term being 1, 1, 2, 2, 3, 3, &c. I write these quantities, one by one, instead of x, not regarding a; and so have - 12= -12, therefore a is not a root; 1+ 3-4 -1+S+4 - 12 8+12-8 12 |