# The product of two consecutive even integers is 168. How do you find the integers?

May 21, 2016

12 and 14
-12 and -14

#### Explanation:

let the first even integer be $x$
So the second consecutive even integer will be $x + 2$
Since the given product is 168 ,the equation will be as follows:

$x \cdot \left(x + 2\right) = 168$

${x}^{2} + 2 \cdot x = 168$

${x}^{2} + 2 \cdot x - 168 = 0$

Your equation is of the form

$a . {x}^{2} + b \cdot x + c = 0$

Find the discriminat $\Delta$

$\Delta = {b}^{2} - 4 \cdot a \cdot c$

$\Delta = {2}^{2} - 4 \cdot 1 \cdot \left(- 168\right)$

$\Delta = 676$

Since $\Delta > 0$ two real roots exist.

$x = \frac{- b + \sqrt{\Delta}}{2 \cdot a}$

$x ' = \frac{- b - \sqrt{\Delta}}{2 \cdot a}$

$x = \frac{- 2 + \sqrt{676}}{2 \cdot 1}$

$x = 12$

$x ' = \frac{- 2 - \sqrt{676}}{2 \cdot 1}$

$x ' = - 14$

Both roots satisfy the condition being even integers

First possibility : two consecutive positive integers

12 and 14

Second possibility : two consecutive negative integers

-12 and -14