# 2 consecutive odd negative integers have a product of 399. What are the integers?

Aug 3, 2016

$- 21$ and $- 19$

#### Explanation:

You know that you're looking for consecutive negative integers, so right from the start you should expect the two two numbers to take the form

$- \left(2 x + 1\right) \to$ the bigger number

$- \left(2 x + 3\right) \to$ the smaller number

This is the case because a positive odd integer can be expressed as

$2 x + 1$

where $x$ is practically any number in $\mathbb{Z}$.

The consecutive even number would be

$\left(2 x + 1\right) + 1 = 2 x + 2$

which makes the consecutive odd number

$\left(2 x + 1\right) + 2 = 2 x + 3$

Since the two numbers are negative, all you have to do is tag along a minus sing.

So, you know that

$- \left(2 x + 1\right) \cdot \left[- \left(2 x + 3\right)\right] = 399$

Expand to get

$4 {x}^{2} + 6 x + 2 x + 3 = 399$

Rearrange to quadratic equation form

$4 {x}^{2} + 8 x - 396 = 0$

Now, this quadratic equation has two possible solutions, as given by the quadratic formula

${x}_{1 , 2} = \frac{- 8 \pm \sqrt{{8}^{2} - 4 \cdot 4 \cdot \left(- 396\right)}}{2 \cdot 4}$

${x}_{1 , 2} = \frac{- 8 \pm \sqrt{6400}}{8}$

${x}_{1 , 2} = \frac{- 8 \pm 80}{8} \implies \left\{\begin{matrix}{x}_{1} = \frac{- 8 + 80}{8} = 9 \\ {x}_{2} = \frac{- 8 - 80}{8} = - 11\end{matrix}\right.$

You need $- \left(2 x + 1\right)$ and $- \left(2 x + 3\right)$ to be negative, which means that your two consecutive odd integers will be

$- \left(2 \cdot 9 + 1\right) = - 19 \text{ }$ and $\text{ } - \left(2 \cdot 9 + 3\right) = - 21$

Do a quick check to make sure that the calculations are correct

$- 19 \cdot \left(- 21\right) = 399 \text{ } \textcolor{g r e e n}{\sqrt{}}$