2d fig. Upon EF (produced if necessary) BA HD Q. E. D. This series of problems might have been extended much further; but the preceding will furnish a sufficient variety, to suggest to the student the best method to be adopted in almost any may occur. The following practical examples are subjoined by way of exercise. Ex. 1. A triangular field, whose sides are 20, 18, and 16 chains, is to have a piece of 4 acres in content fenced off from it, by a right line drawn from the most obtuse angle to the opposite side. Required the length of the dividing line, and its distance from either extremity of the line on which it falls ? Ex. 2. The three sides of a triangle are 5, 12, and 13. If two-thirds of this triangle be cut off by a line drawn parallel to the longest side, it is required to find the length of the dividing line, and the distance of its two extremities from the extremities of the longest side. E.r. 3. It is required to find the length and position of the shortest possible line, which shall divide, into two equal parts, a triangle whose sides are 25, 24, and 7 respectively. 7 Ex, 4. The sides of a triangle are 6, 8, and 10: it is required to cut off nine-sixteenths of it, by a line that shall pass through the centre of its inscribed circle. Er. 5. Two sides of a triangle, which include an angle of 70°, are 14 and 17 respectively. It is required to divide it into three equal parts, by lines drawn parallel to its longest side. Ex. 6. The base of a triangle is 112.65, the vertical angle 57° 57', and the difference of the sides about that angle is 8. It is to be divided into three equal parts, by lines drawn from the angles to meet in a point within the triangle. The lengths of those lines are required. Ex. 7. The legs of a right-angled triangle are 28 and 45. Required the lengths of lines drawn from the middle of the hypothenuse, to divide it into four equal parts. Er. a a a Er. 8. The length and breadth of a rectangle are 1.5 and 9. It is proposed to cut off one-fifth of it, by a line which shall be drawn from a point on the longest side at the distance of 4 from a corner. Ex. 9. A regular hexagon, each of whose sides is 12, is to be divided into four equal parts, by two equal lines; both passing through the centre of the figure. What is the length of those lines when a minimum ? Ex. 10. The three sides of a triangle are 5, 6, and 7. How may it be divided into four equal parts, by two lines which shall cut each other perpendicularly? The student will find that some of these examples will admit of two answers. CHAPTER VIII. ON THE NATURE AND SOLUTION OF EQUATIONS IN GENERAL. 1. In order to investigate the general properties of the higher equations, let there be assumed between an unknown quantity X, and given quantities a, b, c, d, an equation constituted of the continued product of uniform factors: thus (x-a) x (x -b) x (4 -c) x (x-d) = 0. This, by performing the multiplications, and arranging the final product according to the powers or dimensions of x, becomes x4 — a) x3 + ab) x2 - abc x + abcd = 0. (A) -6 b + ac - abd acd d -bcd C + ad + bc + bd + cd Now it is obvious that the assemblage of terms which compose the first side of this equation may become equal to nothing in four different ways; namely, by supposing either x = a, or a = b, or x = 0, or x =d; for in either case one or other of the factors x -2, x-b, x-C, X-d, will be equal to nothing, and nothing multiplied by any quantity whatever will give nothing for the preduct. If any other value e be put for x, then none of the factors e –a, e-b, e-6, 6-d, being equal to nothing, their continued product cannot be equal to nothing. There are therefore, in the proposed equation, four roots 2d fig. Upon EF (produced if necessary) Ex. 1. A triangular field, whose sides are 2 chains, is to have a piece of 4 acres in content fe it, by a right line drawn from the most obtus opposite side. Required the length of the div. its distance from either extremity of the lin falls? This series of problems might have been exter further; but the preceding will furnish a suffici to suggest to the student the best method to be almost any other case that may occur. The foll tical examples are subjoined by way of exercise. Ex. 2. The three sides of a triangle are 5, two-thirds of this triangle be cut off by a lin to the longest side, it is required to find t dividing line, and the distance of its two extr extremities of the longest side. Ex. 3. It is required to find the lengt the shortest possible line, which shall divid parts, a triangle whose sides are 25, 24, an 8 Ex. 4. The sides of a triangle are 6, quired to cut off nine-sixteenths of it, by a through the centre of its inscribed circle. B Ex. 5. Two sides of a triangle, which 70°, are 14 and 17 respectively. It is r into three equal parts, by lines drawn pa side. Let, for example, the enla =-4ac + c2 +α, £ + 20°c + - B2) · (c + d2) = 0. This may of the two subjoined quadratic as contains two imaginary roots ; ing from the product of these two cients of the powers of the unknown Just term of the equation, are real quan ⚫ constituent equations contain imaginary ason is, that these latter disappear by means multiplication. il take place in the equation (x-− a). (z+b). ~2 + d2) = 0, which is formed of two equations degree, and one equation of the second whose ginary. marks being premised, the subsequent general I'll be easily established. THEOREM : roots or values of r; and that which characterizes these roots is, that on substituting each of them successively instead of x', the aggregate of the terms of the equation vanishes by the opposition of the signs + and -; The preceding equation is only of the fourth power or degree; but it is manifest that the above remark applies to equations of higher or lower dimensions : viz, that in general an equation of any degree whatever has as many roots as there are units in the exponent of the highest power of the unknown quantity, and that each root has the property of rendering, by its substitution in place of the unknown quantity, the aggregate of all the terms of the equation equal to nothing. It must be observed that we cannot have all at once x=2, .x = b, x = 0, &c, for the roots of the equation; but that the particular equations x – a = 0, x – b = b = 0, X - C = 0, &c, obtain only in a disjunctive sense. They exist as factors in the same equation, because algebra gives, by one and the same formula, not only the solution of the particular problem from which'that formula may have originated, but also the solution of all problems which have similar conditions. The different roots of the equation satisfy the respective conditions: and those roots may differ from one another, by their quantity, and by their mode of existence. It is true, we say frequently that the roots of an equation are x = a, z = b; x = 0, &c, as though those values of x existed conjunctively; but this manner of speaking is an abbreviation, which it is necessary to understand in the sense explained above. 2. In the equation A, all the roots are positive; but if the factors which constitute the equation had been x + a, l'+b, 2 +6, x + d, the roots would have been negative or subtractive. Thus x4 + a 23 + ab) 22 + abc x + abcd = 0.. (B) + acd it b + bcd tad + bc + bd + cd has negative roots, those roots being x= - a, r = - b, *=-0, x=-d: and here again we are to apply them disjunctively. 3. Some equations have their roots in part positive; in part negative. Such is the following: 73 0 (C) Here ac - bc |