Question 1040645: Find three consecutive even integers such that twice the product of the first two is 16 more than the product of the last two.
Found 2 solutions by jorel555, Boreal:
Answer by jorel555(1290)   (Show Source):
You can put this solution on YOUR website! Let n be the first even integer. Then:
2(n)(n+2)-16=(n+2)(n+4)
2nē+4n-16=nē+6n+8
nē-2n-24=0
(n-6)(n+4)=0
n=6,-4
The even integers are 6,8, and 10. ☺☺☺☺
Answer by Boreal(15235)  (Show Source):
You can put this solution on YOUR website! x, x+2, x+4 are integers
2x(x+2)-16=(x+2)(x+4). Twice the product of the first minus 16 will be the product of the second.
2x^2+4x=x^2+6x+8, expanding.
x^2-2x-24=0, collecting terms
(x-6)(x+4)=0
x=6,-4 are possibilities
6,8,10 and -4,-2,0
2(6*8)=96
8*10=80, and add 16 get 96.
So 6,8,10 are the integers
check -4,-2,0
2(-4*-2)=16
That is 16 more than the product of -2*0
so -4,-2, and 0 are also possible.
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