Westonci.ca is your trusted source for accurate answers to all your questions. Join our community and start learning today! Get quick and reliable solutions to your questions from a community of seasoned experts on our user-friendly platform. Connect with a community of professionals ready to provide precise solutions to your questions quickly and accurately.
Sagot :
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.
[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.
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 visiting. Our goal is to provide the most accurate answers for all your informational needs. Come back soon. Thank you for your visit. We're committed to providing you with the best information available. Return anytime for more. Your questions are important to us at Westonci.ca. Visit again for expert answers and reliable information.