# How do you find two consecutive even integers whose products is 728?

May 11, 2016

Represent two consecutive even integers with a little algebra, then solve the subsequent quadratic to get the two numbers of $26$ and $28$.

#### Explanation:

The first thing we need to do is figure out how to generate an even number.

This is actually a pretty simple task when we think about what an even number is. Consider the first few even numbers:
$2 , 4 , 6 , 8 , 10. . .$

We can see right off the bat that they're all multiples of $2$. And that means to find the $n$th even number (the first, second, third, fourth, etc even number), we simply multiply $n$ (the even number we want to find) by $2$.

For example, if we want to find the 20th even number, we could write down the first 20...
$2 , 4 , 6 , 8 , 10 , 12 , 14 , 16 , 18 , 20 , 22 , 24 , 26 , 28 , 30 , 32 , 34 , 36 , 38 , 40$

...or we could use our neat little formula, $2 n$:
$2 \cdot 20 = 40$

In both cases, we see that the 20th even number is 40.

Now, how about consecutive even numbers? Well, let's start out with some general even number and call it $2 n$. Since each even number is two more than the last, the next even number should be $2 n + 2$. For example, we just found out the 20th even number is 40, by our $2 n$ formula. What's the 21st even number? We just apply our new formula:
$2 \cdot 20 + 2 = 42$

The problem asks us to find the product of two consecutive even integers (and an integer is a fancy word for whole negative/positive number, like $\ldots - 1 , 0 , 1 , 2. . .$). We just found out our two consecutive even numbers to be $2 n$ and $2 n + 2$. The product of these numbers is $\left(2 n\right) \left(2 n + 2\right)$, and we are also told that this must equal 728. And what do you know, we have an equation:
$\left(2 n\right) \left(2 n + 2\right) = 728$
Multiplying out, we have:
$4 {n}^{2} + 4 n = 728$
Subtracting 728 from both sides gives us:
$4 {n}^{2} + 4 n - 728 = 0$

I wouldn't want to factor this equation unless I had to (i.e. my teacher forced me), so using some technology, we find that the roots of this equation are:
$n = - 14$ and $n = 13$

Even though both of these are valid solutions, we'll go with the positive solution of $n = 13$.

So, what does it mean? It means that the 13th even number times the even number after it (the 14th even number) equals 728. Let's first find out what these two numbers actually are, using our neat formulas:
$2 \cdot 13 = 26$
$2 \cdot 13 + 2 = 28$

If you multiply $26$ and $28$, you'll see that it equals $728$. We've found our two consecutive even numbers!