Math question...........?
2009-04-09 3:29 am
Find three positive consecutive even integers such that the product of the first and second is 8 more than 38 times the third.
回答 (4)
2009-04-09 3:41 am
✔ 最佳答案
let x = first integerx + 2 = second integer
x + 4 = third integer
x ( x + 2 ) = 38 ( x + 4 ) + 8
x² + 2x = 38 x + 152 + 8
x² - 36 x - 160 = 0
( x - 40 ) ( x + 4 ) = 0
x = 40
x + 2 = 42
x + 4 = 44
2009-04-09 10:43 am
Let the 3 consecutive even integers be n, n+2, n+4
n(n+2) = (n+4) x 38 + 8
n^2 + 2n = 38n + 152 + 8
n^2 + 2n = 38n + 160
n^2 + 2n -38n -160 = 0
n^2 - 36n - 160 = 0
(n - 40)(n + 4) = 0
n = 40 (reject negative integer)
The answer: 40, 42, 44
n(n+2) = (n+4) x 38 + 8
n^2 + 2n = 38n + 152 + 8
n^2 + 2n = 38n + 160
n^2 + 2n -38n -160 = 0
n^2 - 36n - 160 = 0
(n - 40)(n + 4) = 0
n = 40 (reject negative integer)
The answer: 40, 42, 44
2009-04-09 10:40 am
2 much work 4 me.
2009-04-09 10:38 am
Let the numbers be
(x-2), x and (x+2)
x(x-2) = 38(x+2) + 8
x²-2x = 38x+76+8
x²-40x-84 = 0
(x-42)(x+2) = 0
x = 42
The numbers are {40, 42, 44)
40×42 = 1680
38×44 + 8 = 1680
(x-2), x and (x+2)
x(x-2) = 38(x+2) + 8
x²-2x = 38x+76+8
x²-40x-84 = 0
(x-42)(x+2) = 0
x = 42
The numbers are {40, 42, 44)
40×42 = 1680
38×44 + 8 = 1680
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