although they were presented and solved in a geometrical form, the methods employed have no relation to the generalized conception of algebraic geometry which represents a curve by an equation and vice versa. The simplest quadratic arose in the construction of a mean proportional (x) between two lines (a, b), or in the construction of a square equal to a given rectangle; for we have the proportion a:x = x:b; i.e. x2 = ab. A more general equation, viz. x2 − ax + a2 = 0, is the algebraic equivalent of the problem to divide a line in medial section; this is solved in Euclid, ii. 11. It is possible that Diophantus was in possession of an algebraic solution of quadratics; he recognized, however, only one root, the interpretation of both being first effected by the Hindu Bhaskara. A simple cubic equation was presented in the problem of finding two mean proportionals, x, y, between two lines, one double the other. We have a:x = x:y = y:2a, which gives x2 = ay and xy = 2a2; eliminating y we obtain x3 = 2a3, a simple cubic. The Greeks could not solve this equation, which also arose in the problems of duplicating a cube and trisecting an angle, by the ruler and compasses, but only by mechanical curves such as the cissoid, conchoid and quadratrix. Such solutions were much improved by the Arabs, who also solved both cubics and biquadratics by means of intersecting conics; at the same time, they developed methods, originated by Diophantus and improved by the Hindus, for finding approximate roots of numerical equations by algebraic processes. The algebraic solution of the general cubic and biquadratic was effected in the 16th century by S. Ferro, N. Tartaglia, H. Cardan and L. Ferrari (see Algebra: History). Many fruitless attempts were made to solve algebraically the quintic equation until P. Ruffini and N. H. Abel proved the problem to be impossible; a solution involving elliptic functions has been given by C. Hermite and L. Kronecker, while F. Klein has given another solution.
In the geometric treatment of equations the Greeks and Arabs based their constructions upon certain empirically deduced properties of the curves and figures employed. Knowing various metrical relations, generally expressed as proportions, it was found possible to solve particular equations, but a general method was wanting. This lacuna was not filled until the 17th century, when Descartes discovered the general theory which explained the nature of such solutions, in particular those wherein conics were employed, and, in addition, established the most important facts that every equation represents a geometrical locus, and conversely. To represent equations containing two unknowns, x, y, he chose two axes of reference mutually perpendicular, and measured x along the horizontal axis and y along the vertical. Then by the methods described in the article Geometry: Analytical, he showed that—(1) a linear equation represents a straight line, and (2) a quadratic represents a conic. If the equation be homogeneous or break up into factors, it represents a number of straight lines in the first case, and the loci corresponding to the factors in the second. The solution of simultaneous equations is easily seen to be the values of x, y corresponding to the intersections of the loci. It follows that there is only one value of x, y which satisfies two linear equations, since two lines intersect in one point only; two values which satisfy a linear and quadratic, since a line intersects a conic in two points; and four values which satisfy two quadratics, since two conics intersect in four points. It may happen that the curves do not actually intersect in the theoretical maximum number of points; the principle of continuity (see Geometrical Continuity) shows us that in such cases some of the roots are imaginary. To represent equations involving three unknowns x, y, z, a third axis is introduced, the z-axis, perpendicular to the plane xy and passing through the intersection of the lines x, y. In this notation a linear equation represents a plane, and two linear simultaneous equations represent a line, i.e. the intersection of two planes; a quadratic equation represents a surface of the second degree. In order to graphically consider equations containing only one unknown, it is convenient to equate the terms to y; i.e. if the equation be ƒ(x) = 0, we take y = ƒ(x) and construct this curve on rectangular Cartesian co-ordinates by determining the values of y which correspond to chosen values of x, and describing a curve through the points so obtained. The intersections of the curve with the axis of x gives the real roots of the equation; imaginary roots are obviously not represented.
In this article we shall treat of: (1) Simultaneous equations, (2) indeterminate equations, (3) cubic equations, (4) biquadratic equations, (5) theory of equations. Simple, linear simultaneous and quadratic equations are treated in the article Algebra; for differential equations see Differential Equations.
I. Simultaneous Equations.
Simultaneous equations which involve the second and higher powers of the unknown may be impossible of solution. No general rules can be given, and the solution of any particular problem will largely depend upon the student’s ingenuity. Here we shall only give a few typical examples.
1. Equations which may be reduced to linear equations.—Ex. To solve x(x − a) = yz, y (y − b) = zx, z (z − c) = xy. Multiply the equations by y, z and x respectively, and divide the sum by xyz; then
| (1). |
Multiply by z, x and y, and divide the sum by xyz; then
| (2). |
From (1) and (2) by cross multiplication we obtain
| (suppose) | (3). |
Substituting for x, y and z in x (x − a) = yz we obtain
and therefore x, y and z are known from (3). The same artifice solves the equations x2 − yz = a, y2 − xz = b, z2 − xy = c.
2. Equations which are homogeneous and of the same degree.—These equations can be solved by substituting y = mx. We proceed to explain the method by an example.
Ex. To solve 3x2 + xy + y2 = 15, 31xy − 3x2 − 5y2 = 45. Substituting y = mx in both these equations, and then dividing, we obtain 31m − 3 − 5m2 = 3 (3 + m + m2) or 8m2 − 28m + 12 = 0. The roots of this quadratic are m = 12 or 3, and therefore 2y = x, or y = 3x.
Taking 2y = x and substituting in 3x2 + xy + y2 = 0, we obtain y2 (12 + 2 + 1) = 15; ∴ y2 = 1, which gives y = ±1, x = ±2. Taking the second value, y = 3x, and substituting for y, we obtain x2 (3 + 3 + 9) = 15; ∴ x2 = 1, which gives x = ±1, y = ±3. Therefore the solutions are x = ±2, y = ±1 and x = ±1, y = ±3. Other artifices have to be adopted to solve other forms of simultaneous equations, for which the reader is referred to J. J. Milne, Companion to Weekly Problem Papers.
II. Indeterminate Equations.
1. When the number of unknown quantities exceeds the number of equations, the equations will admit of innumerable solutions, and are therefore said to be indeterminate. Thus if it be required to find two numbers such that their sum be 10, we have two unknown quantities x and y, and only one equation, viz. x + y = 10, which may evidently be satisfied by innumerable different values of x and y, if fractional solutions be admitted. It is, however, usual, in such questions as this, to restrict values of the numbers sought to positive integers, and therefore, in this case, we can have only these nine solutions,
| x = 1, 2, 3, 4, 5, 6, 7, 8, 9; |
which indeed may be reduced to five; for the first four become the same as the last four, by simply changing x into y, and the contrary. This branch of analysis was extensively studied by Diophantus, and is sometimes termed the Diophantine Analysis.
2. Indeterminate problems are of different orders, according to the dimensions of the equation which is obtained after all the unknown quantities but two have been eliminated by means of the given equations. Those of the first order lead always to equations of the form
ax ± by = ±c,
where a, b, c denote given whole numbers, and x, y two numbers to be found, so that both may be integers. That this condition may be fulfilled, it is necessary that the coefficients a, b have no common divisor which is not also a divisor of c; for if a = md and b = me, then ax + by = mdx + mey = c, and dx + ey = c/m; but d, e, x, y are supposed to be whole numbers, therefore c/m is a whole number; hence m must be a divisor of c.
Of the four forms expressed by the equation ax ± by = ±c, it is obvious that ax + by = −c can have no positive integral solutions. Also ax − by = −c is equivalent to by − ax = c, and so we have only to consider the forms ax ± by = c. Before proceeding to the general solution of these equations we will give a numerical example.
To solve 2x + 3y = 25 in positive integers. From the given equation
