# How do you solve x^2 + 6x + 10 = 0 by completing the square?

Apr 3, 2016

${\left(x + 3\right)}^{2} + 1 \to x = \pm i - 3$

#### Explanation:

Completing the square is a method of getting as close as you can with one set of brackets and adding or subtracting the rest.

It ends up in a form like

${x}^{2} + a x + b = {\left(x + c\right)}^{2} + d$

$c$ is always half of $a$, so that when you expand out the brackets you have the right coefficients for ${x}^{2}$ and $x$.

$a = 6 \to c = 3$
${\left(x + 3\right)}^{2} = {x}^{2} + 6 x + 9$

As you can see this is pretty close to the original quadratic. All you need to do is add $1$ and it equates perfectly.

${x}^{2} + 6 x + 10 = {\left(x + 3\right)}^{2} + 1$

To then solve, rearrange like so

${\left(x + 3\right)}^{2} + 1 = 0$
${\left(x + 3\right)}^{2} = - 1$
$x + 3 = \sqrt{- 1} = \pm i$
$x = \pm i - 3$

Apr 3, 2016

There is no solution for the given equation for any $x$ in the set of 'Real Numbers'.

$\textcolor{b l u e}{x = - 3 \pm i}$

#### Explanation:

Consider the standard form of $y = a {x}^{2} + b x + c$

The by completing the square we have:

$y = a {\left(x + \frac{b}{2 a}\right)}^{2} + c - \left[{\left(\frac{b}{2}\right)}^{2}\right]$

In your case $a = 1$ so we have:

$y = {\left(x + \frac{b}{2}\right)}^{2} + c - \left[{\left(\frac{b}{2}\right)}^{2}\right]$

$\implies y = {\left(x + 3\right)}^{2} + 10 - \left({3}^{2}\right)$

$\implies y = {\left(x + 3\right)}^{2} + 1$
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
${x}_{\text{vertex}} = \left(- 1\right) \times \left(+ 3\right) = - 3$
${y}_{\text{vertex}} = + 1$

The coefficient of ${x}^{2}$ = +1 so the graph is of shape type $\cup$

Thus the vertex is a minimum and above the x-axis

Thus there is no solution for $x$ at $y = 0$ where $x \in \mathbb{R}$

However their will be a solution for $x \in \mathbb{C}$ (Complex numbers)

Given that $0 = {\left(x + 3\right)}^{2} + 1$

$\implies \sqrt{{\left(x + 3\right)}^{2}} = \sqrt{- 1}$

$\implies \pm \left(x + 3\right) = i$

Updated: The $\pm$ is for the $x \text{ and } 3$. Not for the $i$

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
color(brown)("Suppose we had "-(x+3)= i

This becomes: $- x - 3 = i$

$\implies - x = + 3 + i$
Multiply by (-1)

$\textcolor{b l u e}{\implies + x = - 3 - i}$
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

$\textcolor{b r o w n}{\text{Suppose we had } + \left(x + 3\right) = i}$

This becomes: $x = - 3 + i$

$\textcolor{b l u e}{\implies + x = - 3 + i}$
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

Solution for $x$ at $y = 0$ is

$\textcolor{m a \ge n t a}{x = - 3 \pm i}$