Find 3 consecutive integers such that the product of the middle and largest integers is five more then the smallest integer squared

Only option: 1,2,3

Step-by-step 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.

credit scoring calculations, such as the fico score, look at a few key factors related to your debt. the amount of overall debt you carry, the ratio of your credit card balances to your credit limit (also called credit utilization), and the relation of your loan balances to the original loan amount.