Question

Question #c3149

Answer 1

`-3` and `-1`
or
`13` and `15`

Explanation:

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

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

product is
`27+6(y)=(2x+1)*(2x+3)`

or

`27+6(2x+1+2x+3)=(2x+1)*(2x+3)`

now we can solve for `x` and plug back into `2x+1` and `2x+3`
for the final answer

`27+24(x+1)=4x^2+8x+3`

`48+24x=4x^2+8x`

`48=4x^2-16x`

`0=x^2-4x-12`

`0=(x+2)(x-6)`

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

Answer 2

`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 `2n`. Even if `n` is odd, `2n` is even. Thus, `2n+1` is guaranteed to be odd.

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

So, we can say that our two consecutive odd integers here are `bb(2n-1)` and `bb(2n+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:

`(2n-1)(2n+1)`

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

`6[(2n-1)+(2n+1)]`

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

`(2n-1)(2n+1)=27+6[(2n-1)+(2n+1)]`

Now we can solve for `n`.

First, add `(2n-1)+(2n+1)`:

`(2n-1)(2n+1)=27+6(4n)`

`(2n-1)(2n+1)=27+24n`

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

`4n^2-1=24n+27`

Move all the terms to the left-hand side:

`4n^2-24n-28=0`

Divide each term by `4`.

`n^2-6n-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:

`(n-7)(n+1)=0`

Implying that:

`n=7` or `n=-1`

We'll take the positive solution of `n=7`. Our two consecutive odd integers are `2n-1` and `2n+1`, so when `n=7`, that translates into `bb13` and `bb15`.

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