How do you determine three consecutive even integers such that the first times the third, is 4 less than 12 times the second?

1 Answer
Apr 21, 2016

-2,0,2
or 10,12,14

Explanation:

First of all, lets call the integers (x-2),(x),(x+2)(x2),(x),(x+2). We can do this because consecutive integers differ by 2. Now from the information we have, we can make an equation:
1st*3rd =12*2nd-41st3rd=122nd4
(x-2)(x+2)=12*(x)-4(x2)(x+2)=12(x)4
x^2-2x+2x-4=12x-4x22x+2x4=12x4
x^2-4=12x-4x24=12x4
x^2=12xx2=12x
x^2-12x=0x212x=0
x(x-12)=0x(x12)=0
Now you see that there are two solutions to this, when x=0x=0 and x=12x=12.
So our integers can be:
-2,0,2
or 10,12,14