# How do you solve using the completing the square method x^2 - 5x = 0?

Feb 1, 2017

I tried this:

#### Explanation:

We can try by adding and subtracting: $\frac{25}{4}$:
${x}^{2} - 5 x + \textcolor{red}{\frac{25}{4} - \frac{25}{4}} = 0$
rearrange:
${x}^{2} - 5 x + \frac{25}{4} = \frac{25}{4}$
Compact it:
${\left(x - \frac{5}{2}\right)}^{2} = \frac{25}{4}$
and...
$x - \frac{5}{2} = = \pm \sqrt{\frac{25}{4}} = \pm \frac{5}{2}$
Two solutions:
${x}_{1} = \frac{5}{2} + \frac{5}{2} = \frac{10}{2} = 5$
${x}_{2} = \frac{5}{2} - \frac{5}{2} = 0$

Feb 1, 2017

$\textcolor{b r o w n}{\text{Solution method given in a lot of detail}}$
The actual process is a lot faster than I have written.

Vertex $\to \left(x , y\right) = \left(\frac{5}{2} , - \frac{25}{4}\right)$
${y}_{\text{intercept}} = 0$

x_("intercpts" )=0" and "5

#### Explanation:

$\textcolor{b r o w n}{\text{It is important that you also look at the general case}}$

Taking it one step at a time: and making it paralleled to the standard form of: $y = a {x}^{2} + b x + c$

The process introduces an error which is canceled out by the inclusion of the correction factor

Let the correction factor be $k$

color(blue)("Step 1 - Group the values ")
$\text{Given: "x^2-5x=0" "color(red)(|)" } a {x}^{2} + b x + c = 0$
Write as $\text{ "(x^2-5x)+k=0 " "color(red)(|)" } a \left({x}^{2} + \frac{b}{a} x\right) + c + k = 0$
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

$\textcolor{b l u e}{\text{Step 2: Take the power outside the bracket}}$

$\text{ "(x-5x)^2+k=0 " "color(red)(|)" } a {\left(x + \frac{b}{a} x\right)}^{2} + c + k = 0$
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
$\textcolor{b l u e}{\text{Step 3: Remove the "x" from } 5 x}$

$\text{ "(x-5)^2+k=0 " "color(red)(|)" } a {\left(x + \frac{b}{a}\right)}^{2} + c + k = 0$

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
$\textcolor{b l u e}{\text{Step 4: Halve the 5}}$

$\text{ "(x-5/2)^2+k=0 " "color(red)(|)" } a {\left(x + \frac{b}{2 a}\right)}^{2} + c + k = 0$

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
$\textcolor{b l u e}{\text{Step 5: Correct the introduced error}}$

The error comes from the color(red)(a)color(green)(xx(+b/(2a))^2 bit

$\text{ "(x-5/2)^2+k=0 " "color(red)(|)" } \textcolor{red}{a} {\left(x \textcolor{g r e e n}{+ \frac{b}{2 a}}\right)}^{\textcolor{g r e e n}{2}} + c + k = 0$

So the correction is:
$\text{ "color(green)(color(red)(1)xx(-5/2)^2)+k=0" "color(red)(|)" } \textcolor{red}{a} \textcolor{g r e e n}{\times {\left(\frac{b}{2 a}\right)}^{2}} \textcolor{b l a c k}{+ k = 0}$

$k + \frac{25}{4} = 0 \text{ "=>" } k = - \frac{25}{4}$

So by substitution we have:

${\left(x - \frac{5}{2}\right)}^{2} + k = 0 \text{ } \to {\left(x - \frac{5}{2}\right)}^{2} - \frac{25}{4} = 0$
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
$\textcolor{b l u e}{\text{Determine key points of the graph}}$

The coefficient of ${x}^{2}$ is +1 so the graph is of general shape $\cup$ thus the vertex is a minimum

${x}_{\text{vertex}} = \left(- 1\right) \times \left(- \frac{5}{2}\right) = + \frac{5}{2}$
${y}_{\text{vertex}} = - \frac{25}{4}$

The general equation of $y = a {x}^{2} + b x + c \to c = 0$
So ${y}_{\text{intercept}} = c = 0$

.......................................................
$\textcolor{b r o w n}{\text{Determine "x_("intercepts}}$

Write as: ${\left(x - \frac{5}{2}\right)}^{2} = \frac{25}{4}$

Square root both sides

$x - \frac{5}{2} = \pm \frac{5}{2}$

$x = \frac{5}{2} \pm \frac{5}{2}$

x_("intercepts")=0" and 5