Answer:
Only option: 1,2,3
Explanation:
3 consecutive integers:
n-1 , n , n+1
Why does it work? They should be consecutive for any integer n.
Example with n=5:
4 , 5 , 6
These are consecutive integers because they happen right after each other in the list of integers.
Integers { ..., -5,-4,-3,-2,-1,0,1,2,3,4,5,...}
We are given the product of the middle and largest is 5 more than the smallest integer squared.
In other words, the product of n and (n+1) is 5 more than (n-1)^2.
As an equation that is: n(n+1) = 5 + (n-1)^2.
Let's write it more together:
n(n+1)=5+(n-1)^2
Distribute:
n^2+n=5+n^2-2n+1 (Used (x-y)^2=x^2-2xy+y^2)
Combine like terms on either side:
n^2+n=n^2-2n+6
Get everything on one side so one side is 0.
Subtract n^2 on both sides:
n=-2n+6
Add 2n on both sides:
3n=6
Divide both sides by 3:
n=2
So if n=2,
then n-1=1
and n+1=3.
So we have one option (1,2,3).
3(2)=5+(1)^2
6=5+1
6=6 is true so (1,2,3) is an option.