Question 283042
The product of two positive consecutive integers is 41 more than their sum. Find the integers.


Let the first number be F


Then the second = F + 1


Their product = F(F + 1) = {{{F^2 + F}}}, and their sum is: F + F + 1 = 2F + 1

 
Since their product is 41 more than their sum, then we have:


{{{F^2 + F = 2F + 1 + 41}}}


{{{F^2 - F - 42 = 0}}}


(F + 6)(F - 7) = 0


Ignore F + 6 = 0, because F = - 6, and we're looking for a positive number. Therefore, F, or the 1st number is 7, and so, the numbers are {{{highlight_green(7_and_8)}}}