Find three consecutive even integers such that the product of the first and second is 8 less than the product of 4 and the third.
2007-06-04 15:54:34 · 7 answers · asked by Anonymous in Science & Mathematics ➔ Mathematics
Let the numbers be 2x, 2x+2, 2x+4.
(2x)(2x+2) + 8 = 4(2x+4)
4x^2 + 4x + 8 = 8x + 16
4x^2 - 4x - 8 = 0
x^2 - x - 2 = 0
(x-2)(x+1) = 0
x = 2, -1
The numbers are: 4, 6, 8 or -2, 0, 2
2007-06-04 16:00:46 · answer #1 · answered by Kemmy 6 · 0⤊ 0⤋
Three consecutive even integers could be represented as x, x+2, and x+4. Thus, we want it so that:
The product of the first and the second ( (x)*(x+2) ) is 8 less than the product of 4 and the third ( = (4*(x+4)) - 8)
Setting this equal gets us:
x(x+2)=4(x+4)-8
x^2 + 2x = 4x + 16 - 8
x^2 - 2x - 8 = 0
(Solve for x, by factoring or quadratic equation)
(x-4)(x+2) = 0
Ao x=4 or x=-2
If x=-2, then x+2 = 0 and x+4=2. To check, is -2 * 0 = 4*2-8 ...yes, so this is a valid solution (-2,0,2)
If x=4, x+2=6 and x+4=8. Check: 4*6 = 4*8-8...yes, so this is also a valid solution (4,6,8)
Thus, this problem has those TWO solutions.
2007-06-04 23:00:19 · answer #2 · answered by Mr. Adkins 4 · 0⤊ 0⤋
Let the integers be x, x + 2, x + 4
x(x + 2) = 4(x + 4) - 8
x^2 + 2x = 4x + 16 - 8
x^2 + 2x = 4x + 8
x^2 - 2x - 8 = 0
(x - 4)(x + 2) = 0
x = 4, -2
x + 2 = 6, 0
x + 4 = 8, 2
We have two sets of integers which satisfy the condition:
(4, 6, 8) and (-2, 0 ,2)
Since you didn't metion anywhere that the integers had to be postive, both the number sets are the solutions.
2007-06-04 23:29:44 · answer #3 · answered by Akilesh - Internet Undertaker 7 · 0⤊ 0⤋
x, x+2 and x+4 are your integers. the first and second times each other are equal to (4* the third) - 8
x(x+2) =4(x+4)-8
2007-06-04 23:01:08 · answer #4 · answered by Jen 3 · 0⤊ 0⤋
x=1st even x+2=2nd even x+4=3rd even #;
x(x+2)=4(x+4)-8 use distributive property
x^2+2x=4x+16-8 Add like terms
x^2+2x=4x+8 add -4x to both sides
x^+2x-4x=4x-4x+8
x^2-2x=8 Now add -8 to both sides
x^-2x-8=8-8
x^2-2x-8=0 Now factor-find 2 # that when you multiply get 8 but when you add give you 2 for middle term
(x-4)(x+2)=0
x-4=0 x+2=0
x=4 x= -2
x+2=6 x+2=0
x+4=8 x+4= 2
2007-06-04 23:03:56 · answer #5 · answered by txmama423 3 · 0⤊ 0⤋
x , x + 2, and x + 4
x(x+2) = 4(x+4) - 8
x^2 + 2x = 4x + 16 - 8
x^2 + 2x = 4x + 8
x^2 - 2x - 8 = 0
(x - 4) (x + 2)
x = 4 or -2
4, 6, 8 or -2, 0, 2
2007-06-04 23:01:52 · answer #6 · answered by 7 · 0⤊ 0⤋
you start with integers x, x+2, x+4.
x(x+2) = 4(x+4) -8
x^2+2x=4x+8
x^2-2x-8=0
(x-4)(x+2)
x=4 x=-2
4,6,8 or
-2,0,2
2007-06-04 23:19:59 · answer #7 · answered by MathGuy 6 · 0⤊ 0⤋