Completing the Square in Vertex Form

Date: 01/13/98 at 19:29:29
From: Carolyn
Subject: Advanced Algebra (Algebra II)

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

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 
solving the quadratic equation:

          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.

This is the famous quadratic formula for solving quadratic equations. 
(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/