# How do you solve 2x^2 - 8x + 16 = 0 by completing the square?

Apr 10, 2016

$x = 2 - 2 i$ or $x = 2 + 2 i$

#### Explanation:

Each monomial in equation $2 {x}^{2} - 8 x + 16 = 0$, is divisible by $2$, so diving by $2$, we get

${x}^{2} - 4 x + 8 = 0$

Now comparing it with ${\left(x - a\right)}^{2} = {x}^{2} - 2 a x + {a}^{2}$, it is apparent that if $- 2 a = - 4$, we have $a = 2$ and ${a}^{2} = 4$.

Hence adding $4$, we can complete the square. Doing so gives us

${x}^{2} - 4 x + 4 - 4 + 8 = 0$ or

${\left(x - 2\right)}^{2} + 4 = 0$ but as in the domain of real numbers the LHS of the equation will be always positive, we cannot have real roots but only complex roots. Hence we write $+ 4$ as #-(2i)^2 and then equation becomes

${\left(x - 2\right)}^{2} - {\left(2 i\right)}^{2} = 0$ or $\left(x - 2 + 2 i\right) \left(x - 2 - 2 i\right) = 0$

i.e. $x = 2 - 2 i$ or $x = 2 + 2 i$