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

Jul 26, 2015

I found: $4 \mathmr{and} 5$ or $- 5 \mathmr{and} - 4$

#### Explanation:

You can write (calling the first integer $n$):
$n \cdot 2 \left(n + 1\right) = 40$
$2 {n}^{2} + 2 n = 40$
so:
$2 {n}^{2} + 2 n - 40 = 0$
${n}_{1 , 2} = \frac{- 2 \pm \sqrt{4 + 320}}{4} = \frac{- 2 \pm \sqrt{324}}{4} = \frac{- 2 \pm 18}{4}$
so:
${n}_{1} = - 5$
${n}_{2} = 4$

Jul 26, 2015

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

#### Explanation:

If consecutive integers, then we are trying to solve:

$n \cdot 2 \left(n + 1\right) = 40$

Divide both sides by $2$ to get:

$n \left(n + 1\right) = 20$

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

$0 = {n}^{2} + n - 20 = \left(n - 4\right) \left(n + 5\right)$

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

$\left(4 , 5\right)$ or $\left(- 5 , - 4\right)$

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

$\left(- 20 , - 1\right)$, $\left(- 10 , - 2\right)$, $\left(- 5 , - 4\right)$, $\left(- 4 , - 5\right)$,
$\left(- 2 , - 10\right)$, $\left(- 1 , - 20\right)$, $\left(1 , 20\right)$, $\left(2 , 10\right)$,
$\left(4 , 5\right)$, $\left(5 , 4\right)$, $\left(10 , 2\right)$, $\left(20 , 1\right)$