x = c, x = T -c = SECTION X.-Equations in General. = a=0, x Equations in general may be prepared or constituted by the multiplication of factors, as we have shown in quadratics. Thus, suppose the values of the unknown quantity x in any equation were to be expressed by a, b, c, d, &c. that is, let x = a, x = b, d, &c. disjunctively, then will æ - b=0, d 0, x 0, &c. be the simple radical equations of which those of the higher orders are composed. Then, as the product of any two of these gives a quadratic equation; so the product of any three of them, as (x — a) (x — b) (x — c) — 0, will give a cubic equation, or one of three dimensions. And the product of four of them will constitute a biquadratic equation, or one of four dimensions; and so on. Therefore, in general, the highest dimension of the unknown quantity x is equal to the number of simple equations that are multiplied together to produce it. = = When any equation equivalent to this biquadratic (aa) (x —b) (x — c) (x — d) = 0 is proposed to be resolved, the whole difficulty consists in finding the simple equations a = 0, x — b = 0, x C = 0, x d =0, by whose multiplication it is produced; for each of these simple equations gives one of the values of x, and one solution of the proposed equation. For, if any of the values of a deduced from those simple equations be substituted in the proposed equation in place of x, then all the terms of that equation will vanish, and the whole be found equal to nothing. Because when it is supposed that x = a, or x = b, or x = = c, or x = d, then the product (x — a) (x — b) (x —c) (x — d) vanishes, because one of the factors is equal to nothing. There are, therefore, four suppositions that give (xa) (xb) (x —c) (x — d) = 0, according to the proposed equation; that is, there are four roots of the proposed equation. And after the same manner any other equation admits of as many solutions as there are simple equations multiplied by one another that produce it, or as many as there are units in the highest dimensions of the unknown quantity in the proposed equation. But as there are no other quantities whatsoever besides these four (a, b, c, d,) that, substituted in the proposed product in the place of x, will make that product vanish: therefore, the equation (a) (x — b) (x —c) (xd) = 0, cannot possibly have more than these four roots, and cannot admit of more solutions than four. - If we substitute in that product a quantity neither equal to a, nor b, nor c, nor d, which suppose e, then since neither, e— b, e -a, e - c, nor e d, is equal to nothing; their product cannot be equal to nothing, but must be some real product: and therefore, there is no supposition besides one of the aforesaid four, that gives a just value of x according to the proposed equation. So that it can have no more than these four roots. And after the same manner it appears, that no equation can have more roots than it contains dimensions of the unknown quantity. To make all this still plainer by an example in numbers, suppose the equation to be resolved to be a 10 x3 + 35 x2 50 x + 24 = 0, and that we discover that this equation is the same with the product of (x-1) (x — 2) (x − 3) (x — 4), then we certainly infer that the four values of x are 1, 2, 3, 4 ; seeing any of these numbers, placed for x, makes that product, and consequently x 10 x3 + 35 x 50 x + 24, equal to nothing, according to the proposed equation. And it is certain that there can be no other values of x besides these four; for when we substitute any other number for x in those factors x−1,x -2, x -3, x -4, none of them vanish, and therefore their product cannot be equal to nothing, according to the equation. A variety of rules, some of them very ingenious, for the solution of equations, may be found in the best writers on Algebra ;* but we shall simply exhibit the easy rule of Trial-andError, as it is given by Dr. Hutton in the 1st volume of his "Course of Mathematics." RULE. "1. Find, by trial, two numbers as near the true root as possible, and substitute them in the given equation instead of the unknown quantity; marking the errors which arise from each of them. "2. Multiply the difference of the two numbers, found by trial, by the least error, and divide the product by the difference of the errors, when they are alike, but by their sum when they are unlike. Or say, as the difference or sum of the errors is to the difference of the two numbers, so is the least error to the correction of its supposed number. * See the treatises of Lacroix, Bonnycastle, Wood, J. R. Young, &c. "3. Add the quotient, last found, to the number belonging to the least error, when that number is too little, but subtract it when too great, and the result will give the true root nearly. "4. Take this root and the nearest of the two former, or any other that may be found nearer; and, by proceeding in like manner, a root will be had still nearer than before; and so on to any degree of exactness required. "Note. It is best to employ always two assumed numbers that shall differ from each other only by unity in the last figure on the right; because then the difference, or multiplier, is 1." Example. To find the root of the cubic equation x3 + x3 + x = 100, or the value of x in it. Here it is soon found that x| Again, suppose 4.2, and 4.3, lies between 4 and 5. Assume, and repeat the work as follows: therefore, these two numbers, and the operation will be as fol lows: Again, suppose 4.264 and 4.265, and work as follows: Then as 064087: 001 :: 027552 : 0.0004299 To this adding 4.264 ... gives a very nearly = 4.2644299 When one of the roots of an equation has been thus found, then take for a dividend the given equation with the known term transposed to the unknown side of the equation; and for a divisor take a minus the root just determined: the quotient will be equal to nothing, and will be a new equation depressed a degree lower than the former. From this a new value of x may be found and so on, till the equation is reduced to a quadratic, of which the roots may be found by the proper rules. SECTION XI.-Progression. When a series of terms proceed according to an assignable order, either from less to greater or from greater to less, by continual equal differences or by successive equal products or quotients, they are said to form a progression. If the quantities proceed by successive equi-differences they are said to be in Arithmetical Progression. But if they proceed in the same continued proportion, or by equal multiplications or divisions, they are said to be in Geometrical Progression. If the terms of a progression successively increase, it is called an ascending progression: if they successively decrease, it is called a descending progression. Thus, 1, 3, 5, 7, 9, &c. form an ascending arithmetical 24, 22, 20, 18, 16, &c. form a descending arithmetical 1, 3, 9, 27, 81, &c. form an ascending geometrical 4, 2, 1,,, &c. form a descending geometrical Progression Arithmetical Progression. 1. Let a be the first term of an arithmetical progression d the common difference of the terms z the last term n the number of terms s the sum of all the terms. Then a, a+d, a+2 d, a + 3 d, &c. is an ascending progression. and a, ad, a -2 d, a-3 d, &c. a descending progression. Hence, in an ascending progression, a + (n-1) d, is the last term; in a descending progression, a (n-1) d, is the last term. 2. Let a series be a + (a + d) + (a + 2 d) + (a+3 d). The same inverted, (a + 3 d)+(a + 2 d) + (a + d) + a. The sum of the two, (2 a + 3 d) + (2 a + 3 d ) + (2 a + 3 d) +(2a + 3 d) = 2 s. That is, (2 a + 3 d) × 4, in this case (a + a + 3 d ) n=2 s. Consequently, s = (a + a = 3 d) n, or= (a + z) n, since 1⁄2 z is here a 3 d. The same would be obtained if the progression were descending; and let the number of terms be what it may. 3. From the equations z=a + (n 1) d, s = in (a + z), and sn (a+a+ (n-1) d), we may readily deduce the following theorems applicable to ascending series. When the series is descending, either the signs of the terms affected with d must be changed, or a must be taken for z; and vice versa. (1.) a = z —nd + d √(− 2 s d + z 2 + d z + 4 d3)+ = 28 } d = 2 + b d — in d = 2 (2.) n 2 2. n n 1 (3.) z = a + nd n + and · d = √(2 s d + a2 — a d + } do) — § d = n a. (4.) s=} n (a+z)=(a+} n d −} d) n=(z—} n d + d) n |