# How do you solve x^2-12x+5=0 by completing the square?

##### 3 Answers
Jan 15, 2018

just add and subtract

#### Explanation:

so we have
${x}^{2} - 12 x + 5 = 0$
${x}^{2} - 12 x + 36 - 36 + 5 = 0$
${\left(x + 6\right)}^{2} - 31 = 0$
therefore
$x + 6 = \pm {31}^{\frac{1}{2}}$
so
$x = \pm {31}^{\frac{1}{2}} - 6$
hope u find it helpful :)

Jan 15, 2018

$x = 6 \pm \sqrt{31}$

#### Explanation:

$\text{using the method of "color(blue)"completing the square}$

• " the coefficient of the "x^2" term must be 1 which it is"

• " add/subtract "(1/2"coefficient of x-term")^2" to"
${x}^{2} - 12 x$

$\Rightarrow {x}^{2} + 2 \left(- 6\right) x \textcolor{red}{+ 36} \textcolor{red}{- 36} + 5 = 0$

$\Rightarrow {\left(x - 6\right)}^{2} - 31 = 0$

$\Rightarrow {\left(x - 6\right)}^{2} = 31$

$\textcolor{b l u e}{\text{take the square root of both sides}}$

$\Rightarrow x - 6 = \pm \sqrt{31} \leftarrow \textcolor{b l u e}{\text{note plus or minus}}$

$\Rightarrow x = 6 \pm \sqrt{31}$

Jan 15, 2018

See explanation

#### Explanation:

For a shortcut method see https://socratic.org/s/aMzZC8RW
This is actually changing

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

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

Where $a {\left(\frac{b}{2 a}\right)}^{2} + k = 0$
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
$\textcolor{b l u e}{\text{The more formal approach}}$

This uses the 'perfect square' but modifies it so that it matches the given equation. A bit like the logic of $\text{something"-x+x="something}$

Perfect square $\to {\left(a + b\right)}^{2} = {a}^{2} + 2 a b + {b}^{2}$

Compare:

${a}^{2} + 2 a b + {b}^{2} = 0 \text{ "..."Standard form}$
${x}^{2} - 12 x + 5 = 0 \text{ "..."Given equation}$

${a}^{2} = {x}^{2} \implies a = x \text{ "......"Point(1)}$

$2 a b \to 2 x b \to - 12 x \implies b = - 6 \text{ "....."Point(2)}$

Using Point(1) and Point(2) now we have:

a^2 +2(a)(color(white)("dd")b)color(white)(".")+color(white)("dd")b^2color(white)("dd")=0" Standard form"
${x}^{2} + 2 \left(x\right) \left(- 6\right) + {\left(- 6\right)}^{2} = 0 \text{ Standard form}$
x^2color(white)("ddd")-12xcolor(white)("dd")+color(white)("dd")5color(white)("d.d")=0" "..."Given equation"

We now have to change the standard form so that it has the same overall value of the given equation. We need to change $+ {\left(- 6\right)}^{2}$ so that it becomes $+ 5. \textcolor{w h i t e}{\text{d}} \to 36 - 31 = + 5$. So the modified standard form is:

$\left[{a}^{2} \textcolor{w h i t e}{} + \textcolor{w h i t e}{\text{ddd")2abcolor(white)("dd.")+color(white)("dd")b^2color(white)("dd}}\right] - 31 = 0$
$\left[{x}^{2} + 2 \left(x\right) \left(- 6\right) + {\left(- 6\right)}^{2}\right] - 31 = 0$

${\left(a + b\right)}^{2} - 31 = 0$
${\left(x \textcolor{red}{- 6}\right)}^{2} \textcolor{g r e e n}{- 31} = 0 \leftarrow \text{ Vertex form (completing the square)}$
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
${x}_{\text{vertex}} = \left(- 1\right) \times \left(\textcolor{red}{- 6}\right) = + 6$
${y}_{\text{vertex}} = \textcolor{g r e e n}{- 31}$

Vertex $\to \left(x , y\right) = \left(6 , - 31\right)$
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

For x-intercepts

$x - 6 = \pm \sqrt{31}$

$x = 6 \pm \sqrt{31} \text{ }$ Exact values

$x = 11.57 \mathmr{and} x = 0.42$
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
From $y = a {x}^{2} + b x + c$ we have:
${y}_{\text{intercept}} \to c = 5$
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ 