# How do you solve k^2+ 8k + 12 = 0 by completing the square?

May 27, 2016

$k = - 6$ or $k = - 2$

#### Explanation:

To solve ${k}^{2} + 8 k + 12 = 0$, by completing the square method

remember ${\left(k + a\right)}^{2} = {k}^{2} + 2 k a + {a}^{2}$

Now here we have ${k}^{2} + 8 k$ in place of ${k}^{2} + 2 k a$, hence $2 a = 8$ or $a = 4$ and we should have ${a}^{2}$ or ${4}^{2} = 16$ in addition to ${k}^{2} + 8 k$ to make it a square.

Hence we can write ${k}^{2} + 8 k + 12 = 0$ as

${k}^{2} + 8 k + 16 - 16 + 12 = 0$

or ${\left(k + 4\right)}^{2} - 4 = 0$

or ${\left(k + 4\right)}^{2} - {2}^{2} = 0$

As now this is in the form ${a}^{2} - {b}^{2}$ which can be factorized as $\left(a + b\right) \left(a - b\right)$, we will have

$\left(k + 4 + 2\right) \left(k + 4 - 2\right) = 0$ i.e, $\left(k + 6\right) \left(k + 2\right) = 0$

Hence either $k + 6 = 0$ or $k + 2 = 0$ or

$k = - 6$ or $k = - 2$