Find two consecutive even integers whose product is 288

Question

 I don't know how to do any of this and I really need all the answers.

 

Answer 1

Let the first integer = x

this means that the next integer must be x+2(it's even)

 

So we know x(x+2) = 288

x^2 + 2x = 288

x^2 + 2x - 288 = 0

(x - 16)(x+18) = 0

x = 16 or -18

x + 2 would be 18 or -16

so 16, 18 is your answer or -18, -16

Answer 2

Two consecutive even integers: 2x and 2(x+1)

 

2x*2(x+1) = 288

2x*(2x+2) = 288

4x² + 4x = 288

4x² + 4x - 288 = 0

 

Divide both sides by 4:

 

x² + x - 72 = 0

 

Factors to:

 

(x+9)(x-8) = 0

x = -9, 8

 

The consecutive integers are:

 

2*(-9) = -18

2*(-9+1) = -16

 

2*(8) = 16

2*(8+1) = 18

 

Check:

(-16)(-18) = 288

(16)(18) = 288

Check!

 

Answer 3

Marcus, this isn't as hard as you think, I promise!

 

Let's let x = the first integer.  Then the 2nd is x + 2, right?

 

The product means we multiply them together, so x(x + 2) = 288

 

x² + 2x = 288.

 

We need to move everything to one side, though.  x² + 2x - 288 = 0

 

Now, we have to factor the left side.  We first multiply the first and third coefficients, 1 x -288 = -288.

 

We need to find two factors of -288 that equal 2 when added together. -4 and 72?  Nope.  -16 and 18?  Yes!

 

We split the -2x.  It doesn't matter whether the -16x or 18x goes first.

 

x² - 16x + 18x -288 = 0

 

Now, we take what's common between the first two terms and second two terms and factor them out.

 

x(x-16) + 18(x-16) = 0

 

See how both parentheses are the same?  Then we can combine what we factored out into their own parentheses.

 

(x+18)(x-16) = 0

 

Either or both on the left have to equal 0 for the right to equal 0.

 

x+18 = 0, so x = -18

 

x-16 = 0, so x = 16

 

So, we have two pairs, 16 and 18 (16 x 18 = 288) and -16 and -18 (-16 x -18 = 288)

 

Hope this helps!