# How do you solve using the completing the square method x^2 - 2x + 1 = 18?

Feb 27, 2016

$\textcolor{b r o w n}{\text{Same thing. Just a different presentation}}$
color(blue)(x_("intercept")~= + 5.24" "or" "-3.24 to 2 decimal places
$\textcolor{b l u e}{{y}_{\text{intercept}} = - 17}$
$\textcolor{b l u e}{\text{vertex "-> (x,y)" "->" } \left(1 , - 18\right)}$

#### Explanation:

Given: $\text{ } {x}^{2} - 2 x + 1 = 18$

Subtract 18 from both sides giving:

$\text{ } {x}^{2} - 2 x - 17 = 0$.........................(1)

'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
This one is more straight forward as the coefficient of ${x}^{2}$ is 1
$\textcolor{b l u e}{\text{Step 1}}$

Write as:

$\left({x}^{2} - 2 x\right) - 17 = 0$
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~
$\textcolor{b l u e}{\text{Step 2}}$
$\textcolor{b r o w n}{\text{At this point we will change the intrinsic value of the equation so it}}$ $\textcolor{b r o w n}{\text{will need to have a correction added later.}}$

Take the power to be outside the brackets

${\left(x - 2 x\right)}^{\textcolor{red}{2}} - 17$

Remove the $x$ from the $- 2 x$ leaving just the $\textcolor{red}{- 2}$

${\left(x \textcolor{red}{- 2}\right)}^{2} - 17$

halve the -2 that is inside the brackets so that you have $\textcolor{red}{- 1}$

${\left(x \textcolor{red}{- 1}\right)}^{2} - 17$

$\textcolor{b r o w n}{\text{Now we add the correction}}$
$\textcolor{g r e e n}{\text{This takes the equation back to its original value so we can once again equate it to zero}}$

Let $\textcolor{red}{k}$ be some constant

${\left(x - 1\right)}^{2} \textcolor{red}{+ k} - 17 = 0$ ................................(2)
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
$\textcolor{b l u e}{\text{Step 3}}$

$\textcolor{b r o w n}{\text{By expanding the brackets and comparing to the original equation}}$$\textcolor{b r o w n}{\text{we can determine the appropriate value of } k}$

Expanding equation (2)

$\cancel{{x}^{2}} - \cancel{2 x} + 1 + k - \cancel{17} \text{ "=" } \cancel{{x}^{2}} - \cancel{2 x} - \cancel{17}$

$k = - 1$ so equation (2) becomes:

$\textcolor{b l u e}{{\left(x - 1\right)}^{2} - 18 = 0}$ .................................(3)

By comparing the values in the vertex format equation you can see how to obtain the vertex coordinates.
Multiply the constant inside the bracket by $\left(- 1\right)$ to get ${x}_{\text{vertex}}$ giving 1 and ${y}_{\text{vertex}}$ can be read directly as -18

$\textcolor{b l u e}{\text{vertex "-> (x,y)" "->" } \left(1 , - 18\right)}$ ;~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
$\textcolor{b l u e}{\text{Step 4}}$

y-intercept at $x = 0$

$\implies {\left(x - 1\right)}^{2} - 18 = y \text{ "->" } {\left(0 - 1\right)}^{2} - 18 = y$

$\textcolor{b l u e}{{y}_{\text{intercept}} = - 17}$

'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
$\textcolor{b l u e}{\text{Step 5}}$

x-intercepts when y=0 so from equation (3)

$\sqrt{{\left(x - 1\right)}^{2}} = \sqrt{18}$

$x - 1 = \pm 3 \sqrt{2}$

$x = 1 \pm 3 \sqrt{2}$

color(blue)(x_("intercept")~= + 5.24" "or" "-3.24 to 2 decimal places