How do you solve the equation x^2+2x+6=0 by completing the square?

Dec 20, 2016

There are no Real solutions to this equation
but see below for method of completing the square
and for Complex value solutions.

Explanation:

Remember that a squared binomial takes the form
$\textcolor{w h i t e}{\text{XXX}} {\left(x = \textcolor{red}{a}\right)}^{2} = {x}^{2} + 2 \textcolor{red}{a} x + {\textcolor{red}{a}}^{2}$

So if ${x}^{2} + 2 x$ are the first 2 terms of an expanded squared binomial:
$\textcolor{w h i t e}{\text{XXX}} \textcolor{red}{a} = 1$
and the third term must be ${\textcolor{red}{a}}^{2} = {1}^{2} = 1$
and the complete (expanded) square needs to be $\textcolor{b l u e}{{x}^{2} + 2 x + 1}$

Rewriting the given equation: ${x}^{2} + 2 x + 6 = 0$
to include this complete (expanded) square:
$\textcolor{w h i t e}{\text{XXX}} \textcolor{b l u e}{{x}^{2} + 2 x + 1} + 5 = 0$

or
$\textcolor{w h i t e}{\text{XXX}} \textcolor{b l u e}{{x}^{2} + 2 x + 1} = - 5$

or
$\textcolor{w h i t e}{\text{XXX")color(blue)(} {\left(x + 1\right)}^{2}} = - 5$

We could note at this point that since any Real number squared is greater than or equal to $0$ there can be no Real solutions.

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If you have advanced to Complex numbers:
$\textcolor{w h i t e}{\text{XXX}} \textcolor{b l u e}{x + 1} = \pm \sqrt{- 5} = \pm i \sqrt{5}$
and
$\textcolor{w h i t e}{\text{XXX}} x = - 1 \pm i \sqrt{5}$