4. Required the factors of x2-5x-6000. 5. Required the quadratic whose roots are 10 and -9. 6. Required the quadratic whose roots are 7. Required the factors of a2+3a-28. and . 8.. Required the factors of 2a2c-3cx2-b2c—acx+ bcx2-a2c. 9. Required the quadratic whose roots are 7 and 1 10. Required the factors of 3400+111x—x2 293 CHAP. XIV.-SOLUTION OF GEOMETRICAL PROBLEMS. ART. 217. A line may be represented by a number, expressing how many linear units the line contains. Thus, the line AB B may be represented by any number, as 7; meaning that the line contains 7 inches, 7 feet, 7 miles, or 7 linear units of any kind. In the same manner, the line AB may be represented by any letter, as a; the letter being understood to denote a number of linear units. Also, if the two adjacent sides of a rectangle be represented by any numbers or letters, the product of those numbers or letters will express the number of superficial units in its area: thus, if a and b represent the adjacent sides, ab will represent the area. Hence, if a be the side of a square, its area will be represented by a2. In the same manner, if a, b, C, represent the number of linear units in the length, breadth, and thickness, respectively, of a solid, the number of its solid units will be expressed by the product abc. Conversely, any single letter may be represented by a line, the product of two letters by a rectangle, and the product of three letters by a solid. A rectangle may, however, be represented by a single letter as a; and if one side of such rectangle be repreAlso if a repre sented by x, then the other will be a sent the area of a square, a will represent the side. Conversely, it is not necessary that every product of two factors should be geometrically represented by a rectangle, and every product of three factors, by a solid for ab may represent a line, which is a times the line b, that is, which contains a times as many linear units; and abc may express a rectangle, which is a times the rectangle bc. And as geometry is only concerned with the three dimensions of length, breadth, and thickness, it is plain, if a product consist of 3+n factors, they cannot all denote lines; and n of them, at least, must be taken as forming a coefficient to the geometrical magnitude expressed by the others. The directions that have been given in Art. 148, for the solution of algebraic problems, are applicable also to the solution of geometrical problems. In order to translate the conditions of a problem into equations, the problem must be supposed to be resolved, and relations sought out between the given and required parts: for the purpose of extending these relations, and discovering new ones, additional lines may be drawn, or figures constructed. For this, however, or for the method of translating the conditions into algebraic language, no general rule can be laid down. PROBLEMS. 1. To divide a straight line, so that the rectangle contained by the whole and one part, shall be equal to the square of the other part. (EUCL. II. 11.) a-x; then, by the question, a (a—x) = x2; whence x2+ ax a2. Hence we have (Art. 145) = Here, if the sign + be taken, a is positive (Art. 213), and less than a ; but if the sign-be taken, it is negative, and greater than a that is to say, the conditions are fulfilled, not only by a point C within the line, as the enunciation contemplated, but also by a point C' on its continuation. The position of the point was supposed to be such that, in order to find it, something was to be taken from AB; whereas the point may also have such a position that, to find it, something must be added to AB (Art. 5,6). To answer this new condition, the question may be easily modified thus: In a given straight line, or its continuation, to find a point such that the rectangle contained by the given line and the segment between one of its extremities and the required point, may be equal to the square of the segment between its other extremity and the same point. If the value of x be put under the form it is plain that we shall have (EUCL. II. 11) 2' If a square be described on EF+EA CF, instead of upon EF-EA = AF, and upon the opposite side from AG, BA produced will meet a side of the square in a point H', which will give AB.BH' AH'2; this construc-. tion answers to the negative value. If we had at first supposed the point to have also the position C', and put BC, or BC' = x, we should have had AC-a-x, AC'=x-a: then (a—x)2=ax, or (x—a)2 = ax, either of which would give 2. Given the difference of the perimeter and diagonal of a square, to find the side and diagonal. Let one side AB = x; then perimeter 4x: also let d = difference of perimeter and diagonal; then BC= 4x-d. Then (EUCL. I. 47) 2x2 (4x-d)2; A C B D (Art. 117). But BC =4x-d=x/2; therefore BC = d (4+ √2) √2 14 d (2√2+1) 7 3. Given the line bisecting the vertical angle of a triangle, and the segments into which it divides the base, to find the sides. Let AD a, BD=b, DC=c, AB=x, AC=y. Then (EUCL. VI. 3) b : c : : x : y,.. cx C D -; whence y с y 4. In a right-angled triangle, given the difference of the |