# Question #c3149

Jul 26, 2016

$- 3$ and $- 1$
or
$13$ and $15$

#### Explanation:

two consecutive odd integers
$2 x + 1$ and $2 x + 3$

the sum is
$2 x + 1 + 2 x + 3 = y$

product is
$27 + 6 \left(y\right) = \left(2 x + 1\right) \cdot \left(2 x + 3\right)$

or

$27 + 6 \left(2 x + 1 + 2 x + 3\right) = \left(2 x + 1\right) \cdot \left(2 x + 3\right)$

now we can solve for $x$ and plug back into $2 x + 1$ and $2 x + 3$

$27 + 24 \left(x + 1\right) = 4 {x}^{2} + 8 x + 3$

$48 + 24 x = 4 {x}^{2} + 8 x$

$48 = 4 {x}^{2} - 16 x$

$0 = {x}^{2} - 4 x - 12$

$0 = \left(x + 2\right) \left(x - 6\right)$

so $x = 6$ or $x = - 2$so lets plug in and check

a) $- 3$ and $- 1$

b) $13$ and $15$

both check so we are done

Jul 26, 2016

$13 , 15$

or

$- 3 , - 1$

#### Explanation:

Let's break this problem down:

We have two consecutive odd integers. How can we express that an two integers are odd and consecutive?

Imagine we had some integer $n$. We don't know if this integer is even or odd, however. But, we can guarantee that we have an even integer if we say $2 n$. Even if $n$ is odd, $2 n$ is even. Thus, $2 n + 1$ is guaranteed to be odd.

If $2 n + 1$ is an odd integer, then we know there will be consecutive odd integers $2$ away in either direction, that is: $\left(2 n + 1\right) - 2 = 2 n - 1$, and $\left(2 n + 1\right) + 2 = 2 n + 3$.

So, we can say that our two consecutive odd integers here are $\boldsymbol{2 n - 1}$ and $\boldsymbol{2 n + 1}$.

Now, we need to set up an equation that represents the statement "their [the odd integers'] product is $27$ more than $6$ times their sum."

Let's start with the integers product. Product means multiplication, so the product of our odd integers is just:

$\left(2 n - 1\right) \left(2 n + 1\right)$

However, this product is $27$ times greater than $6$ times their sum. First write $6$ times their sum, mathematically:

$6 \left[\left(2 n - 1\right) + \left(2 n + 1\right)\right]$

And since the product is $27$ times greater, we can set up the equation as follows:

$\left(2 n - 1\right) \left(2 n + 1\right) = 27 + 6 \left[\left(2 n - 1\right) + \left(2 n + 1\right)\right]$

Now we can solve for $n$.

First, add $\left(2 n - 1\right) + \left(2 n + 1\right)$:

$\left(2 n - 1\right) \left(2 n + 1\right) = 27 + 6 \left(4 n\right)$

$\left(2 n - 1\right) \left(2 n + 1\right) = 27 + 24 n$

Distribute (FOIL) on the left-hand side. Notice that since it is in the form $\left(a + b\right) \left(a - b\right) = {a}^{2} - {b}^{2}$, we get:

$4 {n}^{2} - 1 = 24 n + 27$

Move all the terms to the left-hand side:

$4 {n}^{2} - 24 n - 28 = 0$

Divide each term by $4$.

${n}^{2} - 6 n - 7 = 0$

To factor this, we're looking for two integers whose sum is $- 6$ and product is $- 7$. The integers that fit these criteria are $- 7$ and $1$, so we see the factorization of:

$\left(n - 7\right) \left(n + 1\right) = 0$

Implying that:

$n = 7$ or $n = - 1$

We'll take the positive solution of $n = 7$. Our two consecutive odd integers are $2 n - 1$ and $2 n + 1$, so when $n = 7$, that translates into $\boldsymbol{13}$ and $\boldsymbol{15}$.

If we wish to accept negative answers with $n = - 1$, we see that $2 n - 1 = - 3$ and $2 n + 1 = - 1$.