Paul's Online Notes
Paul's Online Notes
Home / Algebra / Solving Equations and Inequalities / Quadratic Equations - Part II
Show Mobile Notice Show All Notes Hide All Notes
Mobile Notice
You appear to be on a device with a "narrow" screen width (i.e. you are probably on a mobile phone). Due to the nature of the mathematics on this site it is best viewed in landscape mode. If your device is not in landscape mode many of the equations will run off the side of your device (you should be able to scroll/swipe to see them) and some of the menu items will be cut off due to the narrow screen width.

Section 2.6 : Quadratic Equations - Part II

3. Complete the square on the following expression.

\[2{z^2} - 12z\]

Show All Steps Hide All Steps

Start Solution

Remember that prior to completing the square we need a coefficient of one on the squared variable. However, we can’t just “cancel” it since that requires an equation which we don’t have.

Therefore, we need to first factor a 2 out of the expression as follows,

\[2{z^2} - 12z = 2\left( {{z^2} - 6z} \right)\]

We can now proceed with completing the square on the expression inside the parenthesis.

Show Step 2

Next, we’ll need the number we need to add onto the expression inside the parenthesis. We’ll need the coefficient of the \(z\) to do this. The number we need is,

\[{\left( {\frac{{ - 6}}{2}} \right)^2} = {\left( { - 3} \right)^2} = 9\] Show Step 3

To complete the square all we need to do then is add this to the expression inside the parenthesis and factor the result. Doing this gives,

\[\require{color}\require{bbox} \bbox[2pt,border:1px solid black]{{2{z^2} - 12z = 2\left( {{z^2} - 6z \,{\color{Red} + 9}} \right) = 2{{\left( {z - 3} \right)}^2}}}\]

Be careful when the coefficient of the squared term is not a one! In order to get the correct answer to completing the square we must have a coefficient of one on the squared term!