Question

The product of the first and twice the second is 40, what are the two integers?

Answer 1

I found: `4 and 5` or `-5 and-4`

Explanation:

You can write (calling the first integer `n`):
`n*2(n+1)=40`
`2n^2+2n=40`
so:
`2n^2+2n-40=0`
Using the Quadratic Formula:
`n_(1,2)=(-2+-sqrt(4+320))/4=(-2+-sqrt(324))/4=(-2+-18)/4`
so:
`n_1=-5`
`n_2=4`

Answer 2

If consecutive integers then `(4, 5)` or `(-5, -4)`, otherwise any pair of integers whose product is `20` will work.

Explanation:

If consecutive integers, then we are trying to solve:

`n * 2(n+1) = 40`

Divide both sides by `2` to get:

`n(n+1) = 20`

Subtract `20` from both sides and multiply out to get:

`0 = n^2+n-20 = (n-4)(n+5)`

So `n=4` or `n=-5`, meaning that the pairs of consecutive integers are:

`(4, 5)` or `(-5, -4)`

If the integers are not necessarily consecutive, then any integer pair of factors of `20` will work:

`(-20, -1)`, `(-10, -2)`, `(-5, -4)`, `(-4, -5)`,
`(-2, -10)`, `(-1, -20)`, `(1, 20)`, `(2, 10)`,
`(4, 5)`, `(5, 4)`, `(10, 2)`, `(20, 1)`