Question

The square of the first added to twice the second is 5, what are the two integers?

Answer 1

There are an infinite number of solutions, the simplest and only positive integer solutions being 1 and 2.

Explanation:

For any `k in ZZ`

let `m=2k+1`
and `n=2-2k-2k^2`

Then:

`m^2 + 2n`

`= (2k+1)^2+2(2-2k-2k^2)`

`=4k^2+4k+1+4-4k-4k^2 = 5`

Answer 2

If they are supposed to be consecutive integers, then the solution with negatives is the first is `-3` and the second is `-2`.
The positive solution is: first is `1` and second is `2`.

Explanation:

Assuming that these are supposed to be consecutive integers and the lesser integer is the first, then we can use:

first = `n` and second = `n+1`

The square of the first is `n^2` and twicwe the second is `2(n+1)`, so we get the equation:

`n^2+2(n+1) = 5`

(Note that this is not a linear equation. It is quadratic.)

Solve:

`n^2+2(n+1) = 5`
`n^2+2n+2 = 5`
`n^2 +2n-3=0`
`(n+3)(n-1) = 0`

`n+3=0` leads to `n=-3` and `n+1` = -2

If we check the answer, we get `(-3)^2+ 2(-2) = 9+(-4)=5`

`n-1=0` leads to `n=1` and `n+1` = 2

If we check this answer, we get `(1)^2+2(2) = 1+4 =5`