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, x^{2} + 2bx + b^{2 }= (x + b)^{2} x^{2} + 2bx + b^{2 }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:
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 = ax^{2} + 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 = x^{2} + 6x + 2 Add and subtract the square of (1/2) of the coefficient of the linear term
Now, here is a summary of the steps of the process:
EXAMPLE 2: Complete the square on x: y = 2x^{2} + 12x 5 (1) The coefficient of the squared term, x^{2}, must be 1 If the coefficient is not 1 then factor it out of the two variable terms y = 2(x^{2} + 6x)  5 (2) Add and subtract the square of (1/2) of the coefficient of the linear term y = 2(x^{2} + 6x + 3^{2}  3^{2})  5 (3) Separate the perfect square trinomial from the constant terms y = 2(x^{2} + 6x + 3^{2})  3^{2}(2)  5 (4) Factor the perfect square trinomial y = 2(x + 3)^{2}  3^{2}(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.
EXAMPLE 3: Complete the square on t: x = 3t ^{2}+ 3t  1 x = 3( t ^{2} t)  1
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]
Recall that in the above examples we changed the equation of a parabola (y = ax^{2} + 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
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
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
