Question

Two consecutive integers have a product of 6, what are the integers?

Answer 1

let the two concecutive integers be x and x+1
as per question:
x(x+1)=6
solve for x

Answer 2

2(2+1)=6? ... 2 times 3 = 6?

Answer 3

ya its right.........but we take it using a general no....

Answer 4

So the final answer is 2,3?

Answer 5

but there are two sets of solutions

Answer 6

let the two consecutive integres be \( \{n,n+1\}\)\[(n)(n+1)=6\]
\[n^2+n=6\]\[n^2+n-6=0\]
\[(n-2)(n+3)=0\]
\[n=2,-3\]
so
\[ \{n,n+1\}=\{2,2+1\}=\{2,3\}\]or\[\{n,n+1\}=\{-3,-3+1\}=\{-3,-2\}\]