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

2 Answers
Feb 1, 2017

Answer:

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#

Feb 1, 2017

Answer:

#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#