Pete challenges his friend Jill to find two consecutive odd integers that have the following relationship. The product of the integers is 3 times the sum of the integers plus 6. Determine if it is possible to find two such integers.

2n+1, 2n+3 - consecutive odd integers

[tex](2n+1)(2n+3)=3(2n+1+2n+3)+6\\ 4n^2+6n+2n+3=3(4n+4)+6\\ 4n^2+8n+3=12n+12+6\\ 4n^2-4n+15=0\\ \Delta=(-4)^2-4\cdot4\cdot15=16-240=-224\\[/tex]

[tex]\Delta<0\Rightarrow n\in \emptyset [/tex]

It's not possible.

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AL2006 AL2006 · Answer 2 · 2015-03-06T13:09:17-05:00

I don't think it is.

If the first integer is 'x', then the second one is (x+2).
Their product is (x²+2x), and their sum is (2x+2).

You have said that (x²+2x) = 3(2x+2) + 6

   x²+2x = 6x + 6 + 6

   x² - 4x - 12 = 0

      (x - 6) (x + 2) = 0

   x = 6 and x = -2

The numbers are ('6' and '8'), or ('-2' and 0).

I honestly don't know what to make of the second pair.
But the first pair satisfies the description:

The product (48) = 3 x the sum (3 x 14 = 42) plus 6.

So '6' and '8' completely work, except that they're not odd integers.