Answered

Join IDNLearn.com and start exploring the answers to your most pressing questions. Our experts provide prompt and accurate answers to help you make informed decisions on any topic.

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.

Sagot :

Konrad509idn
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.
AL2006idn

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.


Your participation means a lot to us. Keep sharing information and solutions. This community grows thanks to the amazing contributions from members like you. Your questions deserve reliable answers. Thanks for visiting IDNLearn.com, and see you again soon for more helpful information.