Lesson Completing the Square

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This Lesson (Completing the Square) was created by by ccs2011(207) About Me : View Source, Show
About ccs2011:

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%5E2+%2Bbx+%2Bc+=+0
End result in vertex form:
a%28x-h%29%5E2+%2Bk+=+0
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The basis of the completing the square method is the relationship between the squared binomial and its resulting trinomial.
--> %28x%2Ba%29%5E2+=+x%5E2+%2B+2ax+%2B+a%5E2
Working backwards, notice that a is half of 2a.
For example, given x%5E2+%2B8x+%2B+16, we know that half of 8 is 4 and 42 is 16
--> x%5E2+%2B8x+%2B16+=+%28x%2B4%29%5E2
Now lets say you have x%5E2+-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%5E2-2x%2B1+=+%28x-1%29%5E2
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I will demonstrate the entire process step-by-step using the following example:
2x%5E2+%2B+12x+-7+=+0
Step 1: Group the first 2 terms together, separating them from the constant term.
--> %282x%5E2+%2B+12x%29+-7+=+0
Step 2: Factor out leading coefficient, for completing the square to work, the coefficient of x2 must be 1.
--> 2%28x%5E2+%2B6x%29+-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%28x%5E2+%2B+6x+%2B9+-9%29+-+7+=+0
--> 2%28%28x%2B3%29%5E2+-9%29+-+7+=+0
Step 4: Distribute and add constants
--> 2%28x%2B3%29%5E2+-18+-7+=+0
--> 2%28x%2B3%29%5E2+-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%28200%2C300%2C-7%2C1%2C-30%2C10%2C2x%5E2%2B12x-7%29



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