This question is pertaining to college algebra

Question

Find two positive consecutive even integers whose product is 168

Answer 1

Let:

   x  =   first positive, even integer

   x+2 = the consecutive, positive, even integer

 

x(x+2) = 168

x² + 2x - 168 = 0

(x-12)(x+14) = 0   [factor or use quadratic formula]

 

So, either x=12 or x=-14.  Disregard negative (-14).

The two consecutive, positive, even integers are : 12, 14

Checking:  Their product is: 12*14 = 168

Answer 2

I1 = 2x; I2 = 2x+2
(2x)(2x+2) = 168

4x² + 4x = 168

(make equation =0)

4x² + 4x -168 = 0
(divide by 4)
x² + x -42 = 0
(factor)
(x+7)(x-6)=0
(only looking for the even integer)
x = 6
(return to fill in)
I1= 12; I2=14

Go Get Your "A"

Answer 3

Hello, thank you for taking the time to post your question!

The key on this type of question is to use the underlying information in order to set up an expression that we can solve algebraically. On this one we have

x(x + 2) = 168 since it’s saying that we have two consecutive even integers that sum to be 168

from there then you want to solve for the value of “x” algebraically

x^2 + 2x – 168 = 0

(x + 14)(x – 12) = 0

x = 12  OR   x = -14

since the problem specifically says to use positive integers you want to use x = 12 as the first one and then (x + 2) = 14 as the second even integer

I hope that helps you get moving in a better direction on this type of question! Feel free to reach out if you have any additional questions beyond that :)