Answered

Westonci.ca offers quick and accurate answers to your questions. Join our community and get the insights you need today. Discover in-depth solutions to your questions from a wide range of experts on our user-friendly Q&A platform. Discover in-depth answers to your questions from a wide network of professionals on our user-friendly Q&A platform.

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 :

konrad509
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.
AL2006

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.


Thank you for choosing our platform. We're dedicated to providing the best answers for all your questions. Visit us again. We hope you found what you were looking for. Feel free to revisit us for more answers and updated information. We're glad you visited Westonci.ca. Return anytime for updated answers from our knowledgeable team.