Answered

Discover the answers to your questions at Westonci.ca, where experts share their knowledge and insights with you. Experience the convenience of finding accurate answers to your questions from knowledgeable experts on our platform. Get detailed and accurate answers to your questions from a dedicated community of experts on our 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.


Thanks for stopping by. We are committed to providing the best answers for all your questions. See you again soon. We hope you found what you were looking for. Feel free to revisit us for more answers and updated information. Thank you for using Westonci.ca. Come back for more in-depth answers to all your queries.