In the previous section we saw how to complete the square to change a quadratic equation in Standard Form into Vertex Form, if the leading coefficient (a) was one.  In this section we will see how that works when a is NOT one. 

Keep in mind, that the only coefficient that remains the same between Standard and Vertex Form is the leading coefficient. 

Also, keep in mind, that as we perform these algebraic manipulations, each step must yield an equation that is equal to the previous in value.  Just like when factoring, the factored version must be equal to the original un-factored version.  So, when we’re done rewriting in vertex form, we should be able to convert our new vertex form into standard form by following the order of operations.

For example:

What we are going to see here is how to Complete the Square to change from Standard Form, like 2x² – 4x + 7, into Vertex Form, like 2(x – 1)² + 5.  Keep in mind, one could find the vertex using the formula, and then plug that into vertex form, keeping the value of a.  However, the purpose here is to learn to complete the square, as it has many other uses.

Let’s change our equation into Standard Form into Vertex Form, as an example.

Change f(x) = 2x² – 4x + 7 into Vertex Form by Completing the Square

Step 1:  Since we need the coefficient a to be the same in both forms, we will need to divide it out of the first two terms, as is shown below.  At the same time we will leave a space after the bx term, followed by a + sign. 

Note that the original c term is OUTSIDE of the parenthesis.  If we distributed the 2 we would get back the original equation.

Step 2:  Now we take half of b – squared, add it, only for now, inside the parenthesis.

What is the 2, outside of the parenthesis, doing to everything inside the parenthesis?  Multiplying, right?  So, even though we only wrote a single + (-1)², it is being multiplied by two.  In order to keep the new equation equal to the original, we must now subtract 2 of these…so, add the – 2(-1)² after the original c term (7).

Remember, we do NOT want to square the b/2 that we added, but we do square the “extra.”

Step 3:  Factor the square, simplify the extra.

Boom, done!

Let’s see an uglier example, one that doesn’t reduce so nicely, in shorter steps.

This seems tricky, but if you understand (1) each step’s purpose, and (2) why each step maintains equality with the original equation, the entire process boils down to just arithmetic.  With practice it gets a lot easier. 

Let’s see one more example.

Step 1:  Divide a out of ax² + bx, leaving space and moving c outside the parenthesis.

This seems tricky, but if you understand (1) each step’s purpose, and (2) why each step maintains equality with the original equation, the entire process boils down to just arithmetic.  With practice it gets a lot easier. 

Let’s see one more example.

Step 1:  Divide a out of ax² + bx, leaving space and moving c outside the parenthesis.

Step 2:  Add inside the parenthesis, then subtract  outside the parenthesis.

Step 3:  Factor the square, simplify the “extra.”

That’s all there is to it.  The fractions work out easy if you’re neat an organized, and the factoring is easy because the value of   – h (from vertex form), is (b/2a). 

Below is a review of the practice problems.

This second video is a live YouTube feed where we reviewed this with students in a “virtual classroom setting.”