# How do you find 2 consecutive even integers whose product is 224?

Mar 27, 2016

Two consecutive even integers are $\left\{14 , 16\right\}$ or $\left\{- 16 , - 14\right\}$

#### Explanation:

Two consecutive even integers could be $n$ and $n + 2$ and as their product is $224$, we have

$n \left(n + 2\right) = 224$ or ${n}^{2} + 2 n = 224$ or

${n}^{2} + 2 n - 224 = 0$

hence $n = \frac{- 2 \pm \sqrt{{2}^{2} - 4 \cdot 1 \cdot \left(- 224\right)}}{2 \cdot 1}$ or

$n = \frac{- 2 \pm \sqrt{4 + 896}}{2 \cdot 1} = \frac{- 2 \pm \sqrt{900}}{2} = \frac{- 2 \pm 30}{2}$

Hence $n = \frac{- 2 + 30}{2} = 14$ or $n = \frac{- 2 - 30}{2} = - 16$

Hence two consecutive even integers are $\left\{14 , 16\right\}$ or $\left\{- 16 , - 14\right\}$

Mar 27, 2016

$\left(14 , 16\right) \mathmr{and} \left(- 14 , - 16\right)$

#### Explanation:

Let the first integer may be $n$

Remember that even numbers differ in $2$

So,the second number will be $n + 2$

color(purple)( :.n(n+2)=224

Use distributive property color(brown)(a(b+c)=ab+ac

$\rightarrow {n}^{2} + 2 n = 224$

$\rightarrow {n}^{2} + 2 n - 224 = 0$

This is a Quadratic equation (in form $a {x}^{2} + b x + c = 0$)

color(brown)(x=(-b+-sqrt(b^2-4ac))/(2a)

Where

color(red)(a=1,b=2,c=-224

$\rightarrow x = \frac{- 2 \pm \sqrt{{2}^{2} - 4 \left(1\right) \left(- 224\right)}}{2 \left(1\right)}$

$\rightarrow x = \frac{- 2 \pm \sqrt{4 - \left(- 896\right)}}{2}$

$\rightarrow x = \frac{- 2 \pm \sqrt{4 + 896}}{2}$

$\rightarrow x = \frac{- 2 \pm \sqrt{900}}{2}$

$\rightarrow x = \frac{- 2 \pm 30}{2}$

Now we have two solutions

color(indigo)((-2+30)/(2)=28/2=14

color(violet)((-2-30)/(2)=-32/2=-16

$n$ is expressed here as $x$

$\therefore n = \left(14 \mathmr{and} - 16\right)$

So, the integers are color(green)((14,16)and(-14,-16)