40. Given a3 — y3: (x − y)3 :: 61 : 1), to find the values of and xy=320 .. (Alg. 180.) 3x2y 3y* : ( − y) :: 60 : 1, or зxy × (x − y): (x − y)3 :: 60: 1; . (Alg. 184.) 960 (xy): 60: 1, dividing the first and second terms by x 2 y; and 16 (xy) :: 1: 1, dividing the first and third terms by 60; From the first equation, x3 + x2y + xy2 + y3 = 2336; and from the second, x3 — x2y — xy2 + y3 .. by subtraction, adding this to the first equation, = 576; 2x2y + 2xy2 = 1760; Dividing the second equation by xy, x+y=4; :: x3 + 3x2y + 3xy2 + y3 = 64. But from the first equation, - x2y — xy2 + y3 = 0; Again, dividing the first equation by this last, Dividing the second equation by the first, (x2 + y2) × (x + y) = 15xy, or x + xy + xy2 + y3 = 15xy; but from the first, xxyxy' + y3 = зxy; SECTION IV. Solution of Adfected Quadratics, involving only one unknown Quantity. (28.) LET the terms be arranged on one side of the equation, according to the dimensions of the unknown quantity, beginning with the highest; and (17) the known quantities be transposed to the other side; then, if the square of the unknown quantity has any coefficient, either positive or negative, let all the terms be divided by this coefficient (13). If the square of half the coefficient of the second term be now added (11) to both sides of the equation*, that side which involves the unknown quantity will become a complete square; and (19) extracting the square root on both sides of the equation, a simple equation will be obtained, from which the values of the unknown quantity may be determined. * This is called completing the square; and that a complete square is thus obtained may be easily proved. Let x2+2ax be the proposed quantity on one side, when the terms are arranged according to the form prescribed above; and suppose y2 = the quantity requisite to complete the square. Now the square of x±d = x2±2dx + d2, where it is evident that four times the product of the extreme terms is equal to the square of the middle term; and therefore, in order that x2ax + y2 may be a square, 42 y2 must be equal to 442x2; therefore y2=a2 the square of half the coefficient of the middle term. |