Asked by Jessy
Find four consecutive integers, such that the product of the first and second integer minus 2 is equal to 3 times the product of the third and fourth integer.
Answers
Answered by
Steve
let the numbers be x,x+1,x+2,x+3
x(x+1)-2 = 3(x+2)(x+3)
x = -2 or -5
check:
-2(-1)-2 = 0 = 3(0)(1)
-5(-4)-2 = 18 = 3(-3)(-2)
x(x+1)-2 = 3(x+2)(x+3)
x = -2 or -5
check:
-2(-1)-2 = 0 = 3(0)(1)
-5(-4)-2 = 18 = 3(-3)(-2)
Answered by
Reiny
let the 4 consecutive intergers be
x-1, x , x+1 and x+2
(x-1)(x) -2 = 3(x+1)(x+2)
x^2 - x - 2 = 3x^2 + 9x + 6
2x^2 + 10x + 8 = 0
x^2 + 5x + 4 = 0
(x+1)(x+4) = 0
x = -1 or x = -4
case 1: the numbers are -2, -1, 0, and 1
or
case 2: the numbers are -5, -4, -3, and -2
x-1, x , x+1 and x+2
(x-1)(x) -2 = 3(x+1)(x+2)
x^2 - x - 2 = 3x^2 + 9x + 6
2x^2 + 10x + 8 = 0
x^2 + 5x + 4 = 0
(x+1)(x+4) = 0
x = -1 or x = -4
case 1: the numbers are -2, -1, 0, and 1
or
case 2: the numbers are -5, -4, -3, and -2
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