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,

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

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**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+8n-28=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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