A student sits studying a book in a brightly lighted library

When you are asked to complete the square, it means that they want you to go from the standard form of
f(x) = ax2 + bx +c
to the vertex form of the equation
f(x) = a(x − h)2 + k, where (h, k) is the vertex of the parabola.

You can download a handwritten pdf of this page here: Completing the Square. 

Shortcut Method

  1. Note that “a” is the same for both equations.
  2. Find the vertex, (h, k)
  3. Substitute a, h, and k into f(x) = a(x − h)2 + k

And that’s it – you are done!

Long, Tedious Method

Used when you need to show all the steps. This is easier explained with an example, so I will walk through completing the square using this quadratic equation:

f(x) = 2x2 − 12x + 14

Step 1:

Put parentheses around the quadratic and linear terms, then factor “a” out of the parentheses.

\large \mathrm{f(x) = ( \color{red}2 \color{black}x^2 \color{green} - 12 \color{black}x) \color{purple} + 14} \\ \: \\ \large \mathrm{f(x) = \color{red}2 \color{black}(x^2 \color{green} - \dfrac{12}{\color{red}2} \color{black} x) \color{purple} + 14} \\ \: \\ \large \mathrm{f(x) = \color{red}2 \color{black}(x^2 \color{green} - 6 \color{black} x) \color{purple} + 14}

Step 2:

Take 1/2 of the linear coefficient. 

\large \mathrm{\dfrac{1}{2} \cdot ( \color{green} -6 \color{black} ) = \color{green} -3}

Step 3:

Square it.

\large \mathrm{( \color{green} -3 \color{black} )^2 = \color{green} +9}

Step 4:

Add it to the terms inside of the parentheses, but to keep everything equal, subtract it on outside of the parentheses.

Watch out! There’s a sneaky bit here. Because a (in our example, “2“) is multiplying everything inside of the parentheses, we need to multiply our result from step 2 by a before subtracting it! You are basically adding and subtracting the same amount to your equation, which is the equivalent of adding a fancy zero so everything is still the same but looks different. You’ll see why in a minute.

\large \mathrm{f(x) = \color{red}2 \color{black}(x^2 \color{green} - 6 \color{black} x \color{purple} + 9 \color{black} ) \color{purple} + 14 \color{black} - \color{red} 2 \color{black} \cdot \color{purple} 9 } \\ \: \\ \large \mathrm{f(x) = \color{red}2 \color{black}(x^2 \color{green} - 6 \color{black} x \color{purple} + 9 \color{black} ) \color{purple} + 14 \color{black} - \color{purple} 18 } \\ \: \\ \large \mathrm{f(x) = \color{red}2 \color{black}(x^2 \color{green} - 6 \color{black} x \color{purple} + 9 \color{black} ) \color{magenta} - 4 }

Step 5:

Factor the new quadratic that is inside of the parentheses. It is now a perfect square.

The new term outside of the parentheses is k, the y-value of the vertex. (That is why it switched color.)

\large \mathrm{f(x) = \color{red}2 \color{black}(x - 3)(x - 3) \color{magenta} - 4 }
\\ \: \\ \large \boxed{ \mathrm{f(x) = \color{red}2 \color{black}(x - \color{darkorange} 3 \color{black} )^2 \color{magenta} - 4}} \\ \: \\

And we’re done! We finished completing the square, and this is the vertex form, with a vertex of (3, -4).

Check our work with the Shortcut Method

  1. Note that “a” is the same for both equations: a = 2
  2. Find the vertex, (h, k)

    h = −b/(2a)   =  −(−12)/(2·2)   =   12/4   =   3

    k = f(3) = 2(32) − 12(3) + 14 = −4

  3. Substitute a, h, and k into the vertex form gives us: f(x) = 2(x − 3)2 + (−4)

Our work checks out – we got the same answer with both methods.

It probably seems like a lot to remember right now, but the key to this becoming easier is practice. Eventually, it will be easier and faster.

Questions? Thoughts about this? Another topic you would like to see covered? Please leave a comment below!