Mesa Community College


College Algebra Concepts - MAT 150 online

Completing the Square
Instructor: Dr. Jo Steig

 

Completing the square is a process that can be used to solve an equation, find the vertex on a parabola, or to put an equation of a conic in standard form. On this page I am going to give several examples of completing the square that will allow us to find the vertex of a parabola.

BACKGROUND: The basic idea behind completing the square is that we are going to build an expression that will factor into a perfect square. Note that in the following,

x2 + 2bx + b2 = (x + b)2

x2 + 2bx + b2 is called a perfect square trinomial because it factors into (x + b)2, a squared binomial. In the process of completing the square we will build a perfect square trinomial that will factor into a squared binomial.

Note that in the above example the following is true:

(1) The coefficient of the squared term, x2, is 1

(2) The second term of (x + b)2 , b, is (1/2) of the coefficient of the linear term of the trinomial, x2 + 2bx + b2. That is,

The linear term of x2 + 2bx + b2 is 2bx

The coefficient of the linear term is 2b

(1/2) of 2b = b, the second term of (x + b)2

 We will use this information in the process of completing the square.

 

First, I will give and example of completing the square and will then follow that with detailed steps of the process. In the following examples we will change the equation of a parabola (y = ax2 + bx + c) into the form y = a( x - h) 2 + k. Determining the vertex from this form will be discussed at the end of the discussion of completing the square.

 

EXAMPLE 1: Complete the square on x: y = x2 + 6x + 2

Add and subtract the square of (1/2) of the coefficient of the linear term

The linear term is 6x

The coefficient of the linear term is 6

(1/2) of the coeffient of the linear term is (1/2) 6 = 3

The square of (1/2) of the coeffient of the linear term is 32

y = x2 + 6x + 9 - 9+ 2

y = (x2 + 6x + 9) - 9+ 2

y = (x + 3)2 - 7

 

Now, here is a summary of the steps of the process:

Completing the Square

(1) The coefficient of the squared term, x2, must be 1
If the coefficient is not 1 then factor it out of the two variable terms (example will follow)

(2) Add and subtract the square of (1/2) of the coefficient of the linear term

(3) Separate the perfect square trinomial from the constant terms

(4) Factor the perfect square trinomial

(5) Add the constant terms

 

EXAMPLE 2: Complete the square on x: y = 2x2 + 12x- 5

(1) The coefficient of the squared term, x2, must be 1

If the coefficient is not 1 then factor it out of the two variable terms

y = 2(x2 + 6x) - 5

(2) Add and subtract the square of (1/2) of the coefficient of the linear term

y = 2(x2 + 6x + 32 - 32) - 5

(3) Separate the perfect square trinomial from the constant terms

y = 2(x2 + 6x + 32) - 32(2) - 5

(4) Factor the perfect square trinomial

y = 2(x + 3)2 - 32(2) - 5

(5) Add the constant terms

y = 2(x + 3)2 - 23

 

You should take particular notice that in step (3) above, the constant term that came out of the parentheses had to be multiplied by 2.

Here is another example, but without the commentary:

 

EXAMPLE 3: Complete the square on t: x = -3t 2+ 3t - 1

x = -3( t 2- t) - 1

x = -3( t 2- t + .5 2 - .5 2 ) - 1

x = -3( t 2- t + .5 2 ) - .5 2 (-3) - 1

x = -3( t - .5 )2 - .5 2 (-3) - 1

x = -3( t - .5 )2 + .75 - 1

x = -3( t - .5 )2 - .25

 

This process can be used even with numbers that would be considered unwiedy without a calculator - you just follow the same steps and pattern shown above. Here is such an example.

 

EXAMPLE 4: Complete the square on x: y = 42.94x 2+ 62.34x - 12.31

y = 42.94 (x 2+ 1.45x) - 12.31 [1.45 = 62.34/42.94]

y = 42.94 (x 2+ 1.45x + .73 2 - .73 2 ) - 12.31[.73 = 1.45/2]

y = 42.94 (x 2+ 1.45x + .73 2 ) - .73 2 (42.94) - 12.31

y = 42.94 (x + .73 ) 2 - .73 2 (42.94) - 12.31

y = 42.94 (x + .73 ) 2 - 22.88 - 12.31

y = 42.94 (x + .73 ) 2 - 35.19

 

DETERMINE THE VERTEX OF A PARABOLA

 

Recall that in the above examples we changed the equation of a parabola (y = ax2 + bx + c) into the form y = a( x - h) 2 + k.In this latter form the vertex is identified as the point with coordinates (h,k). Here is why:

A parabola that opens upward or downward is either going to have a minimum value of y (if the parabola opens upward) or a maximum value of y (if the parabola opens downward).
If a > 0 then the parabola opens upward and there is a minimum value for y. This will occur when a( x - h) 2 = 0 for that is the smallest value that a (x - h) 2 can take on.
a( x - h) 2 = 0

( x - h) 2 = 0

( x - h) = 0

x = h, so the x-coordinate of the vertex is h.

When x = h, y = k so the coordinates of the vertex are (h,k)

 

 
If a < 0 then the parabola opens downward and there is a maximum value for y. This will occur when a( x - h) 2 = 0 for that is the largest value that a(x - h) 2 can take on (recall that a < 0).
a ( x - h) 2 = 0

( x - h) 2 = 0

( x - h) = 0

x = h, so the x-coordinate of the vertex is h.

When x = h, y = k so the coordinates of the vertex are (h,k)

 

Now let's go back to each of the previous four examples and identify the vertex.

FROM EXAMPLE 1: y = (x + 3)2 - 7

Vertex ( - 3, - 7)

FROM EXAMPLE 2: y = 2(x + 3)2 - 23

Vertex ( - 3, - 23)

FROM EXAMPLE 3: x = -3( t - .5 )2 - .25

Vertex (.5, - . 25) [assuming that the ordered pairs are in the form, (t,x)]

FROM EXAMPLE 4: y = 42.94 (x + .73 ) 2 - 35.19

Vertex: ( - .73, - 35.19)

 

 © 2000 Jo Steig