Question

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

Answer 1

I tried this:

Explanation:

We can try by adding and subtracting: `25/4`:
`x^2-5x+color(red)(25/4-25/4)=0`
rearrange:
`x^2-5x+25/4=25/4`
Compact it:
`(x-5/2)^2=25/4`
and...
`x-5/2==+-sqrt(25/4)=+-5/2`
Two solutions:
`x_1=5/2+5/2=10/2=5`
`x_2=5/2-5/2=0`

Answer 2

`color(brown)("Solution method given in a lot of detail")`
The actual process is a lot faster than I have written.

Vertex `->(x,y)=(5/2,-25/4)`
`y_("intercept")=0`

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

Explanation:

`color(brown)("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=ax^2+bx+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 ")`
`"Given: "x^2-5x=0" "color(red)(|)" "ax^2+bx+c=0`
Write as `" "(x^2-5x)+k=0 " "color(red)(|)" "a(x^2+b/ax)+c+k=0`
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

`color(blue)("Step 2: Take the power outside the bracket")`

`" "(x-5x)^2+k=0 " "color(red)(|)" "a(x+b/ax)^2+c+k=0`
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
`color(blue)("Step 3: Remove the "x" from "5x)`

`" "(x-5)^2+k=0 " "color(red)(|)" "a(x+b/a)^2+c+k=0`

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
`color(blue)("Step 4: Halve the 5")`

`" "(x-5/2)^2+k=0 " "color(red)(|)" "a(x+b/(2a))^2+c+k=0`

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
`color(blue)("Step 5: Correct the introduced error")`

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

`" "(x-5/2)^2+k=0 " "color(red)(|)" "color(red)(a)(xcolor(green)(+b/(2a)))^(color(green)(2))+c+k=0`

So the correction is:
`" "color(green)(color(red)(1)xx(-5/2)^2)+k=0" "color(red)(|)" "color(red)(a)color(green)(xx(b/(2a))^2)color(black)(+k=0)`

`k+25/4=0" "=>" "k=-25/4`

So by substitution we have:

`(x-5/2)^2+k=0" "->(x-5/2)^2-25/4=0`
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
`color(blue)("Determine key points of the graph")`

The coefficient of `x^2` is +1 so the graph is of general shape `uu` thus the vertex is a minimum

`x_("vertex")=(-1)xx(-5/2)=+5/2`
`y_("vertex")=-25/4`

The general equation of `y=ax^2+bx+c -> c=0`
So `y_("intercept")=c=0`

.......................................................
`color(brown)("Determine "x_("intercepts")`

Write as: `(x-5/2)^2=25/4`

Square root both sides

`x-5/2=+-5/2`

`x=5/2+-5/2`

`x_("intercepts")=0" and 5`