Learning intentions:
In this section we will examine:
Completing the square (CTS)
The process of completing the square (CTS) allows us to convert a quadratic in the general form (y = ax2 + bx + c) into turning point form (y = a(x - h)2 + k). In the turning point the turning point (vertex) is located at (h, k).
Method: Completing the square
Remember, to complete the square the coefficient of x2 must be 1. If the coefficient is any number other than 1, simply divide all terms by the coefficient:
Once the coefficient of x2 is 1, use the following steps to complete the square:
The expression will now been in turning point form:
4F - VIDEO EXAMPLE 1:
Factorise the following quadratic by completing the square:
4F - VIDEO EXAMPLE 2:
Factorise the following quadratic by completing the square:
4F - VIDEO EXAMPLE 3:
Factorise the following quadratic by completing the square:
Solving quadratic equations by completing the square
Once a you have completed the square on a quadratic, it is possible to find the solutions to the equation by using inverse operations to isolate x.
4F - VIDEO EXAMPLE 4:
Solve the following quadratic for x, given it is already in turning point form:
4F - VIDEO EXAMPLE 5:
Solve the following quadratic for x, by first completing the square:
The axis of symmetry
Completing the square on the general quadratic formula
Consider the general quadratic equation:
We can complete the square on the general form:
Therefore, after completing the square on the general quadratic formula the expression is:
The axis of symmetry
From the equation above we can see that the turning point (vertex) will exist upon the axis of symmetry defined by:
Where the values of a and b come from the general form: y= ax2 + bx + c.
Figure 1 - The axis of symmetry lies half-way between the two x-intercepts.
4F - VIDEO EXAMPLE 6:
Use the axis of symmetry to determine the turning point for the following quadratic. Hence, determine the turning point form without completing the square.
Dynamic geometry representation of the axis of symmetry
The following GeoGebra gadget shows the axis of symmetry as the x-intercepts (m,0) and (n,0) are changed. As you can see the turning point (vertex) always exists on the line of symmetry for the parabola regardless of the value of the x-intercepts.
Success criteria:
You will be successful if you can: