119k views
4 votes
Find three positive consecutive integers such that the product of the first and second is 2 more than 9 times the third

1 Answer

4 votes
Three positive consecutive integers are n, n+1, and n+2. And these integers satisfy:

n(n+1)=9(n+2)+2 expanding each side gives you:

n^2+n=9n+18+2

n^2+n=9n+20 subtract 9n from both sides

n^2-8n=20 subtract 20 from both sides

n^2-8n-20=0 now factor

n^2+2n-10n-20=0

n(n+2)-10(n+2)=0

(n-10)(n+2)=0, since we only want positive integers

n=10

So the three integers are 10, 11, 12
User Cyclomarc
by
8.1k points

No related questions found

Welcome to QAmmunity.org, where you can ask questions and receive answers from other members of our community.

9.4m questions

12.2m answers

Categories