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

Answer:

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

Explanation:

First of all, lets call the integers #(x-2),(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-4#
#(x-2)(x+2)=12*(x)-4#
#x^2-2x+2x-4=12x-4#
#x^2-4=12x-4#
#x^2=12x#
#x^2-12x=0#
#x(x-12)=0#
Now you see that there are two solutions to this, when #x=0# and #x=12#.
So our integers can be:
-2,0,2
or 10,12,14