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### Completing the Square in Vertex Form

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Date: 01/13/98 at 19:29:29
From: Carolyn

Dear Dr. Math,

I am a sophmore taking Advanced Algebra. I am having trouble with
these kinds of problems like this:

y = 10x squared + 10x + 1

We are supposed to put it into vertex form (y-k = a(x-h)squared.
I sort of get the process but I always make a different mistake,
and even more important, I need to know why every step is done.

I would appreciate it if your answer was in laymen's terms and work is
shown.

Thank you for your time. It is greatly appreciated.

Sincerely,
Carolyn
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Date: 01/26/98 at 11:02:57
From: Doctor Joe
Subject: Re: Advanced Algebra (Algebra II)

Dear Carolyn,

I used to get the wrong answer for this type of problem too, but
here's how I got over it:

Let a, b, c be constants.

Suppose we are looking at the expression y = ax^2 + bx + c.

Step 1:  Isolate the c from the ax^2 + bx term

y = {ax^2 + bx} + c

Step 2:  Next, pluck out the a:

y = a{x^2 + b/a} + c

Step 3:  Add the term a(b/2a)^2 to both sides.

y + a(b/2a)^2 = a(x^2 + b/a + (b/2a)^2) + c

Step 4:  Complete the square.

y + b^2/4a = a(x+b/2a)^2 + c

Step 5:  Bring the constant over to the y-side.

y - (c - b^2/4a) = a(x + b/2a)^2

This is the required form.

Now, by putting a = 10, b = 10, c = 1, you should be able to work out
the right answr.

The reason why you need this form is that it is extremely useful for

ax^2 + bx + c = 0

Here we set y = 0 in the above "vertex" form (commonly known as the
completing the square),

-(c- b^2/4a) = a(x + b/2a)^2

Now, we have

(x + b/2a)^2 = (b^2 - 4ac)/4a^2

So, we have

x + b/2a = + sqrt{(b^2 - 4ac)/4a^2} or
- sqrt{(b^2 - 4ac)/4a^2}.

It follows that

x = {-b + sqrt(b^2 - 4ac)}/2a or
{-b - sqrt(b^2 - 4ac)}/2a.

(QED)

P.S.  Ever wonder why the form is called the "vertex" form?

-Doctor Joe,  The Math Forum
Check out our web site!  http://mathforum.org/dr.math/
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Associated Topics:
High School Basic Algebra

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