Lesson Completing the Square

Source code of 'Completing the Square'

This Lesson (Completing the Square) was created by by ccs2011(207)

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The purpose of this lesson is to demonstrate how to write a quadratic equation in vertex form using the completing the square method.
Given a standard quadratic equation: 
{{{ax^2 +bx +c = 0}}}
End result in vertex form:
{{{a(x-h)^2 +k = 0}}}

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The basis of the completing the square method is the relationship between the squared binomial and its resulting trinomial.
--> {{{(x+a)^2 = x^2 + 2ax + a^2}}}
Working backwards, notice that a is half of 2a.
For example, given {{{x^2 +8x + 16}}}, we know that half of 8 is 4 and 42 is 16
--> {{{x^2 +8x +16 = (x+4)^2}}}
Now lets say you have {{{x^2 -2x}}} what do i need to add to make it a perfect square?
Well half of (-2) is (-1) and (-1)2 is 1. Therefore, by adding 1 it becomes a perfect square.
--> {{{x^2-2x+1 = (x-1)^2}}}

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I will demonstrate the entire process step-by-step using the following example:

{{{2x^2 + 12x -7 = 0}}}

Step 1: Group the first 2 terms together, separating them from the constant term.

--> {{{(2x^2 + 12x) -7 = 0}}}

Step 2: Factor out leading coefficient, for completing the square to work, the coefficient of x2 must be 1.

--> {{{2(x^2 +6x) -7=0}}}

Step 3: Complete the square, Take half of x coefficient and square it. Notice to keep equation balanced you must add this number and subtract it making the net effect zero.

--> {{{2(x^2 + 6x +9 -9) - 7 = 0}}}

--> {{{2((x+3)^2 -9) - 7 = 0}}}

Step 4: Distribute and add constants

--> {{{2(x+3)^2 -18 -7 = 0}}}

--> {{{2(x+3)^2 -25 = 0}}}

Now it is successfully in vertex form and can be easily graphed.
The vertex is at (-3,-25)
The parabola opens up and has a y-intercept at (0, -7)
Here is a graph of this parabola:

{{{graph(200,300,-7,1,-30,10,2x^2+12x-7)}}}