CONSTRUCTION of Equations, nic sections may cut one another, in Algebra, is the method of find-j can be determined, will arise to ing the roots of equations by means four dimensions, and the conic of geometrical figures, which is sections may be so assumed as to effected by lines or curves, accord- make this equation coincide with ing to the order or degree of the any proposed biquadratic; so that equation. the ordinates from these four intersections will represent the four roots of the proposed biquadrate. To construct a Simple Equation. -This is done by resolving the equation into a proportion, thus: the general form for simple equations being ax =b, or ax = 1. b, it is only necessary to find a fourth proportional to the three quantities, a, 1, and b; which fourth proportional will give the value of r. To construct a Quadratic Equation. If it be a simple quadratic, it will reduce to this form 2 ab; whence ux: x: b, or x=ab, a mean proportional between a and b. Let M M/ M" M (plate II. fig.7,) be a parabola whose axis is AP, and M M M M a circle whose centre is C and radius C M, cutting the parabola in the points M M M" M; and from these points draw the ordinates to the axis MP, M' P', MP, MP, and from C draw CD perpendicular to the axis, and CN parallel to it meeting PM in N. Let AD-a, DC=b, CM = n, the parameter of the parabola=p, AP=x, PMy; then pr=y2. Also CM2 CN2 + NM2, or n2= (x-a)2 + (y − b )2; Adfected quadratics will be of three forms, according to the signs of the terms, the general equation 22axb, not admitting of a construction that will apply to alln2, the varieties. or x2-2ux a2 + y2 — 2by+b2 and substituting for x its value , 1. If the quadratic ber2+2ax and arranging the terms acb; then form the right-angled triP angle whose base AB (plate II. fig. 5,) is a, and perpendicular BC is b; with the centre A and radius AC, describe the semi-circle DCE; then DB and BE will be the two roots of the given quadratic; being = √(b2+ a2) ± a. 2. If the quadratic be 22-2ax b2, the construction will be the same as that of the preceding one, x= a + √(b2+a2) having one positive and one negative value. cording to the powers of y, this If one of the intersections of the conic section fall upon the axis, then one of the ordinates vanishes, and the equation by which these ordinates are determined will then be only of three dimensions, or a 3. But if the form be 2-2ax cubic, to which any proposed cu-b2, or 2ax-x2b2; construct bic equation may be accommoa right-angled triangle whose hy-dated, so that the three remaining pothenuse (plate II. fig. 6,) FG is ordinates will be the three roots of a, and perpendicular GH is bthe proposed cubic. with the radius FG and centre F, CONSTRUCTION of Geometrical describe a semicircle IGK; then Problems. shall IH and HK be the roots of the given equation, a being a (a2 — b2). In this form, if a be greater than b, the equation will have two positive roots; but if a be less than b, the solution is manifestly impossible. To construct Cubic and Biquadratic Equations.-The roots of a biquadratic equation may be determined by the intersection of two conic sections; for the equation by which the ordinates, from the four points in which these co PROB. 1. Giving the base of a right-angled triangle, and the dif ference between the hypothenuse and perpendicular, to construct the triangle. Const. Let A B (plate II. fig. 8,) be the given base, and BD the difference of the other two sides. From the extremity B, of the given base A B, demit the perpendicular BD, equal to the given difference, and produce it indefinitely towards C. Join AD, and from the point A draw A C, ma' ing the angle DAC ADC, and meeting BD produced in C; so shall ABC be the triangle required. Demons. Since the angle DAC ADC, the side ACCD (Euc. vi. 1), therefore the difference between AC and CB is equal to BD, but BD is equal to the given difference by construction, and AB is equal to the given base; also the angle B is a right angle: consequently, ABC is the right-angled triangle required. which are such, that the one of them multiplied into the fluxion of the other, may be of the form of the fluxion proposed; then taking the fluxion of the proposed rectangle, there will be deduced a value of the original fluxion, in terms that will frequently admit of finite fluents. This will be better illustrated by an example. 1. Let it be proposed to find the x2 x fluent of. ing y = √(22 + a2), we shall If the difference BD is equal to, or exceeds the base AB, the prob lem is impossible; for in either of those cases the line AC would not meet the continuation of BD, which is necessary for the construction. have one part of the fluxion of This limitation arises from a known this product; that is, 2 x F√(x2 geometrical problem, though it is+a2) of the same form as the pronot found in Euclid's "Elements;" posed fluxion. viz. "The difference of any two sides of a triangle is less than the third side." PROB. 4. Giving the hypothenuse of a right-angled triangle, and the side of its inscribed square, to construct the triangle. Now putting the assumed rec. tangle into fluxions, we have or or x2 + a2 √(x2 + u2) y= Whence Construc. On the given hypothenuse AB (Plate II. fig. 9) describe the circle ACD, and from the centre G demit the perpendicular GD; join BD, and perpendicular to it, at the point B, draw BE = to half the diagonal of the given square; join DE, and produce it to F, making EF BE. Then from D as a centre, and with DF as a radins, describe the arc FC cutting now the fluent of the circle in C; join AC, BC, and ABC will be the triangle required. log. x + √(x2+a2) CONTACT, Angle of, is that made by a curve line and a tangent to it, at the point of contact, is less than any right-lined angle whatever; √(x2 + a2) — § a2 though it does not therefore follow that it is of no magnitude or quan. tity. CONTINUAL Proportionals are a series of quantities in which the ratio between each two adjacent terms is equal. √(x2+a2) = 3 ≈ hyp. log. x + √(x2+ a2); } e a CONTINUED Fractions, ar ex certain species of fractions tremely useful in various arithmetical and indeterminate problems, in the reduction of ratios, the exCONTINUATION, in the Flux-traction of roots, &c. ional Analysis, or the finding of Every fraction of the form Fluents by Continuation, is a method of finding one fluent from another. For from which it is deduced, is ra- nominator for the first fraction. tional or irrational. For the second fraction, multiply The above is the most general the terms of the first by the se form that can be given to continu-cond denominator, adding 1 to the ed fractions, but there are few product of the denominator. cases in which it is necessary to every succeeding one multiply the have any others than those whose terms of the last by the corres numerators are unity, and signs+, ponding denominator, and add to and therefore it will be sufficient the products the terms of the last to consider the latter form only. except one. The series of fractions formed of the first term, the first two terms, the first three terms, &c. of any continued fraction, are called converging fractions. To reduce any proposed rational Fraction to a continued Fraction. Divide the denominator by the numerator, then the divisor by the remainder, and so on; and the successive quotients will be the several denominators of the continued fraction, 1 being the From this it follows that the dif ference of any two contiguous fractions is 1 divided by the product of their denominators, and that the difference between the product of the numerator of each, by the other's denominator, is also 1. To extract the Square Root by continued Fractions. Since the quantity that we have to represent by a continued frac tion is irrational, the fraction which expresses it must consist of an indefinite number of terms; but still it has this property, that after a certain period, the denominators of the several fractions recur again in the same order as at first; and consequently, after having arrived at that period, the series may be carried on to any length at pleasure. Let it be required to extract the square root of 19 by continued fractions; we first find the greatest integer contained in √19, which is 4; therefore, 19-4 or multiplying √19=4+ 171 1 9748 8+1 this last fraction by √19+4 we √19++ 19- 16 =4++ √19 +4 &c. and in the same way we may |