Posted by Stupid on Thursday, October 9, 2008 at 9:37pm.
Find three consecutive even integers such that twice the product of the first and last numbers is 28 more than the square of the second number.
I know the answers i just need to know how to get to them and ive spent 30 min on it HELP ME!!!!

math  bobpursley, Thursday, October 9, 2008 at 9:43pm
let the integers be n, n+1, and n+2
2n*(n+2)28=(n+1)^2
solve for n, notice it is a quadratic.

math  Stupid, Thursday, October 9, 2008 at 9:45pm
Buy why wouldn't it be n, n+2, and N+4 because it is even ingergers,

math  Not answered, Thursday, October 9, 2008 at 9:56pm
this question is still not answered will some one walk through it step by step

math  Reiny, Thursday, October 9, 2008 at 10:51pm
you are right, it should have been
n, n+2 and n+4 so let's adjust bobpursley's equation to look like
2n(n+4)  28 = (n+2)^2

math  Adithya, Friday, October 17, 2008 at 7:40pm
lets start with the equation
2n(n+4)  28 = (n+2)^2
using formula (a+b)^2=a^2+2ab+b^2
2n^2+8n28=n^2+4n+32
simplyfying,
n2+4n=32
Solving for n, we get n=4
So the even consecutive numbers are 4,6,and 8.....
checking...
2(4*8)28=36
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