Question

The product of two consecutive even integers is 48. Which equation can be used to find these integers?

Answer 1

Since they are both even, let x and x + 2 be the integers.
x(x + 2) = 48
x² + 2x - 48 = 0
(x - 6)(x + 8) = 0

So, at x = 6 and x = -8.
So you can have 6 and 8 or -8 and -6 as your two integers.

Answer 2

Two consecutive even integers are 2n and 2(n+1), where n = any integer. So, the equation we will use is this:

----------------------------------->        $n(n+2)=48$        <--------------------------------------

Since the integers are even and are consecutive, we know that if we add two to an even number, the next even number will be its consecutive integer. That is why we multiply n by itself when 2 is added with it. Let's find out what our integers are.

$n^{2} +2n = 48$........................Here we distributed the n across the parenthesis

$n^{2} +2n-48=0$......................Subtract 48 from both sides to get it into standard quadratic form

$(n+8)*(n-6) = 0$........................Here we factored by asking ourselves which two numbers add to equal 2 but multiply to equal -48 ( 8 and -6)

Now we set our parenthesis equal to 0 and find out what our n can equal.

$(n+8) = 0$
$n = -8$

$(n-6)=0$
$n=6$

Now we found out two possible forms of n. Now if we plug it into our equation, we will notice that for each of the numbers we plug in, we multiply it by itself added to 2. So:

$6*(6+2)=48$
$6*(8)=48$

And,

$-8*(-8+2)=48$
$-8*(-6)=48$

These equations are both true, so n can equal -8 or -6.

Thank you for the question!! I hope this helped! Have an amazing day! :D