CHAPTER XIX SIMULTANEOUS QUADRATIC EQUATIONS 332. The degree of an equation involving several unknown quantities is equal to the greatest sum of the exponents of the unknown quantities contained in any term. xy+y=4 is of the second degree. x3y + 5 x2y3 — y4 is of the fifth degree. 333. A symmetrical equation is one which is not altered by interchanging the unknown quantities. x + y = xy, x2 + x1y+ + y2 = 4 are symmetrical equations. x - y = 2 and x3 — y3 = 1 are not symmetrical, but a change of sign would make them symmetrical. 334. A homogeneous equation is an equation all of whose terms are of the same degree with respect to the unknown quantities. 4x3-3x2y 3 y3 and x2 - 2 xy - 5 y2 = 0 are homogeneous equations. 335. The absolute term of an equation is the term which does not contain any unknown quantity. In x2-4 xy + 2 = 0 the absolute term is 2. 336. Simultaneous quadratic equations involving two unknown quantities lead, in general, to equations of the fourth degree. A few cases, however, can be solved by the methods of quadratics. * *The graphic solution of simultaneous quadratic and higher equations has been treated in Chapter XVII. I. EQUATIONS SOLVED BY FINDING x + y AND x − y 337. If two of the quantities x+y, x-y, xy are given, the third one can be found by means of the relation (x + y)2 — 4 xy = (x-y)2. 338. In many cases two of the quantities x+y, x − y, and xy are not given, but can be found. 339. The roots of simultaneous quadratic equations must be arranged in pairs, e.g. the answers of the last example are: II. ONE EQUATION LINEAR, THE OTHER QUADRATIC 340. A system of simultaneous equations, one linear and one quadratic, can be solved by eliminating one of the unknown quantities by means of substitution. III. HOMOGENEOUS EQUATIONS 341. If one equation of two simultaneous quadratics is homogeneous, the example can always be reduced to an example of the preceding type, for one unknown quantity can be expressed in terms of the other. Consider the homogeneous equation, 4x2-11 xy+6 y2=0. (1) Expressing x in terms of y by means of the formula for quadratics, x= 11 y± √(11 y)2 - 4 · 4 · 6 y2 8 11 y± 5 y =2y, or y. In most cases this result can be obtained more simply by factoring, e.g. factor (1), Combining these results with another quadratic equation produces two systems of the preceding kind. |