# How do you solve x^2 + 10x + 13 = 0 by completing the square?

May 15, 2016

$\text{ "x~~ -8.464" and } - 1.536$

#### Explanation:

Given: $y = {x}^{2} + 10 x + 13 = 0$

This process introduces an error that has to be compensated for. To do this I introduce a corrective as yet unknown value represented by $k$. The value of $k$ may be determined after all the other changes have taken place

Compare to $y = a {x}^{2} + b x + c$

This then written as a first step as:

$y = a \left({x}^{2} + \frac{b}{a} x\right) + c$ in your case $a = 1$

$\textcolor{b l u e}{\text{Step 1}}$

$y = \left({x}^{2} + 10 x\right) + 13 + {k}_{1} \leftarrow \text{ At this point } {k}_{1} = 0$

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$\textcolor{b l u e}{\text{Step 2}}$

Move the square from ${x}^{2}$ outside the brackets. This begins to introduce errors.

$y = {\left(x + 10 x\right)}^{2} + 13 + {k}_{2}$

$\textcolor{b r o w n}{\underline{\text{Just for Evan}}}$
color(brown)(k_2" is whatever is needed to turn "y=(x+10x)^2+13
color(brown)("back to "y=x^2+10x+13

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$\textcolor{b l u e}{\text{Step 3}}$

Remove the $x$ from the $10 x$

$y = {\left(x + 10\right)}^{2} + 13 + {k}_{3}$

$\textcolor{b r o w n}{\underline{\text{Just for Evan}}}$
$\textcolor{b r o w n}{{k}_{3} \text{ is whatever is needed to turn } y = {\left(x + 10\right)}^{2} + 13}$
color(brown)("back to "y=x^2+10x+13

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
$\textcolor{b l u e}{\text{Step 4}}$

Halve the 10

$y = {\left(x + 5\right)}^{2} + 13 + {k}_{4}$

$\textcolor{b r o w n}{\underline{\text{Just for Evan}}}$
$\textcolor{b r o w n}{{k}_{4} \text{ is whatever is needed to turn } y = {\left(x + 5\right)}^{2} + 13}$
color(brown)("back to "y=x^2+10x+13
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NOW WE DETERMINE THE VALUE OF $k$

Just for demonstration lets multiply out the brackets and then compare what we have to the original equation.

$y = {x}^{2} + 10 x \textcolor{red}{+ 25} + 13 + {k}_{4} \leftarrow \text{ New equation}$
y=x^2+10x color(white)("dd.d")+13 color(white)("ddd.")larr" Original equation"

For the new equation to work we must have $25 + {k}_{4} = 0 \implies \textcolor{p u r p \le}{{k}_{4} = - 25}$
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$\textcolor{b l u e}{\text{Step 5 : Substitute for } {k}_{4}}$

$\textcolor{g r e e n}{y = {\left(x + 5\right)}^{2} + 13 \textcolor{p u r p \le}{+ {k}_{4}} \textcolor{w h i t e}{\text{ddd") -> color(white)("ddd}} y = {\left(x + 5\right)}^{2} + 13 \textcolor{p u r p \le}{- 25}}$

color(green)(color(white)("dddddddddddddddddddd") ->color(white)("dd")y=(x+5)^2-12
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
$\textcolor{b l u e}{\text{Now we solve for } x}$

Set $\text{ } y = 0 = {\left(x + 5\right)}^{2} - 12$

$12 = {\left(x + 5\right)}^{2}$

Square root both sides

$\pm \sqrt{12} = x + 5$

Subtract 5 from both sides

$x = - 5 \pm 2 \sqrt{3}$

Jan 1, 2018

$x = - 5 \pm 2 \sqrt{3}$

#### Explanation:

Start with ${x}^{2} + 10 x + 13 = 0$

Subtract 13 from both sides:

${x}^{2} + 10 x = - 13$

Take 10, divide by 2, to get 5. Square 5 to get 25. Add 25 to both sides:

${x}^{2} + 10 x + 25 = - 13 + 25$

${x}^{2} + 10 x + 25 = 12$

The left side is now a perfect square:

${\left(x + 5\right)}^{2} = 12$

Now solve by taking the square root of both sides:

$x + 5 = \pm \sqrt{12}$

Subtract 5 from both sides:

$x = - 5 \pm \sqrt{12}$

Simplify the radical if you'd like:

$x = - 5 \pm 2 \sqrt{3}$