QUADRATIC EQUATIONS. 2a = 7 and b = 641, we shall get 5 and 2 for the sought numbers. 15. If 2a and b are the sum and the sum of the fifth powers of two numbers or quantities, it is required to find them. Representing them by a + and a-x, we must have (a + x)3 + (a − x) = b, or 2a + 20a3x2 + 10ax1 = b; which b 2a5 gives a1 + 2a2x2 = 10a Solving the equation, we have x = 1b + 8a5 - a2); 1b + 8a5 10a — a2) /b+8a 10a a2), consequently, a + and a as required. If 2a = 6 and b 1056, we get 4 and 2 for the sought numbers. = 16. A person on a journey goes the distance a the first day, ad the second day, a + 2d the third day, and so on, in arithmetical progression; then how long will it take him to go the distance b? Ans. 2α d √(4a2 — 4ad +d2+8bd) days. 2d + 2d 17. To find two positive members, such that if the square of the first is multiplied by 5 and that of the second by 7, the sum of the products shall equal 157; and that if the square of the first is multiplied by 7 and that of the second by 5, the sum of the products shall equal 143. Ans. 3 and 4. 18. Having the sum, 3a, and the sum of the squares, b, of three numbers or quantities, in arithmetical progression, to find the numbers or quantities. Representing the progression by xy, x, and x+y, the conditions of the question give the equations xy + x + x + y = 3a or x = a, and (x − y)2+ 19. Given, the sum, 4a, and the sum of the squares, b, of four numbers or quantities, in arithmetical progression, to find the numbers or quantities. Representing the progression by x − 3y, x − y, x + y, and +3y, the conditions give the equations 4x=4a or x=a, and (x−3y)2+(x− y)2+(x+y)2+(x+3y)2=4x2+20y2 = 4a2+ -; consequently, the terms of the 20y=b, or y = b- 4a2 20 progression are easily found. = 24 and b = 20 = 164, we have a = 6 and y = 1; consequently, the progression is 3, 5, 7, 9. Remark.-If we have the sum, na, and sum of the squares or cubes or fourth powers b, of n numbers or quantities in arithmetical progression, then we can, by methods similar to those used in this and the preceding question, find the terms of the progression from the solution of quadratic equations. 20. Given, the sum, 3a, and the product, b, of three numbers or quantities, in arithmetical progression, to find the numbers or quantities. - Representing the progression by x − y, x, and ∞ + Y, x the conditions give x = a and (x − y)x(x + y) = x(x2 — y2) = = Remark. If we have the sum and product of four or five numbers or quantities, in arithmetical progression, then it is easy to perceive (from what has been done) how the progression may be found by quadratics. 21. "Having the sum a, and the sum of the squares b, of four numbers or quantities, in geometrical progression, to find the progression." Let x and y denote the second and third terms; then, by the nature of such a progression, y3 and must be the first y x and fourth terms; consequently, from the conditions, we + x2 + y2+ ข x xy x2 + y2 α xy 1), it is clear that the first of the preceding equa and it is easy, in like manner, to show that the second is equivalent to -2; consequently, by substitution, the 2 x2 + y2 α a and 2= = xy x + y + y Eliminating + y2 from the preceding equations, and adding 2 to each member of the first, we shall get (4 tions, etc.) we have (x + y) + (x + y) = Solving this quadratic, we get a +y=√√/@2 b 2a = ; consequently, if we substitute the value of x + y in xy (x + y)3 a + 2(x + y) we shall find the value of xy. Hence, from the values of x + y and xy we can easily find the values of x and y, and thence the terms of the progression become known; thus, if a,15, and 685, we get x+y=6 and xy = 8, and solving these equations, we have x2 x = 2 and y = 4, and thence the progression T, x, y, and y becomes 1, 2, 4, and 8. x 22. Having the sum, a, and the sum of the squares, b, of five numbers or quantities, in geometrical progression, to find the progression. Let x and y denote the second and third terms; then, by Hence, by the question, we must have the equations first, fourth, and fifth terms. x y1 yo + x + y + + =a and + 2x2 + y2 + + b; and dividing the second of these equations by the first, we tion of the question is reduced to that of the equations By taking the half sum and half difference of these equa multiplying the terms of the second of these by, and sub tracting the result from the first, we shall have Multiplying the terms of the second of these equations by 2a +y, which is easily reduced to the quadratic y2 Solving the quadratic, and taking the positive root, we (5a6a2b+56) ; consequently, having found y, we easily get x from the equation x + x a2 - b 2a ; thus, if a 31 and b = 341, we have y = 4 and x = 2, and thence the progression is 1, 2, 4, 8, and 16. 23. Having a, the sum of the means, and b, the sum of the extremes, of four numbers or quantities, in geometrical progression, to find the progression. Ans. Supposing the terms of the progression to increase from left to right, then a equals the difference of a b+3a the means, and of course the progression is easily found. Thus, if a 24 and 656, the progression is 2, 6, 18, and 54. 24. Having a, the sum of the extremes, and b, the product, of three numbers or quantities, in harmonical progression, to find the progression. Let x, y, and z denote the progression; then, since their reciprocals are in arithmetical progression, we have the |