Question

Note: This is NOT a question. This is a tutorial.

How to use completing the square method? See comment below to see how!

Answer 1

First, we must familiarize ourselves with the general equation.

\[ax^2 + bx + c = 0\]

this is the general equation

The first step in completing the square method is to get rid of the coefficient of \(x^2\). By that, I mean turning the coefficient of \(x^2\) into 1. How do we do that? By dividing EVERY term of the equation by a (which is the coefficient of x).

\[\Large \frac{ax^2}{a} + \frac{bx}{a} + \frac{c}{a} = 0\]
a gets canceled out...

\[\Large \frac{\cancel{a}x^2}{\cancel{a}} + \frac{b}{a} x + \frac{c}{a} = 0\]
Now, we isolate x by "transposing" \(\frac{c}{a}\) to the other side of the equation. Transposing means to put a term into the other side of the equation then changing the signs. Others prefer the phrase "adding by the additive inverse". The additive inverse is the negative of that term. In our case, \(\frac{c}{a}\) is the term so the "additive inverse" is \(-\frac{c}{a}\). So, we add that to our equation.

\[\Large x^2 + \frac{b}{a} x + \cancel{\frac{c}{a}} - \cancel{\frac{c}{a}} = 0 - \frac{c}{a}\]
\[\Large x^2 + \frac{b}{a} x = -\frac{c}{a}\]
Now, what do we do? We will asign a \(\mathit{new}\) third term for our equation. One that will make our left hand side a perfect square trinomial. How to find for the third term? First, we divide the COEFFICIENT of x (which in this case is \(\frac{b}{a}\)) by 2. So we have \(\frac{b}{2a}\). Then what? Now, we square the result. So, we have \((\frac{b}{2a})^2 = \frac{b^2}{4a^2}\). We add this to our equation...

\[\Large x^2 + \frac{b}{a} x + \frac{b^2}{4a^2} = -\frac{c}{a}\]
but wait! Since we added something on the left hand side, to maintin equality, we shall add the same value to the right hand side so we have..

\[\Large x^2 + \frac{b}{a} x + \frac{b^2}{4a^2} = -\frac{c}{a} + \frac{b^2}{4a^2}\]
we can simplify this further! Since our right hand side has an LCD of \(4a^2\), we will add them...


\[\Large x^2 + \frac{b}{a} x + \frac{b^2}{4a^2} = \frac{-4ac + b^2}{4a^2}\]
Now, as we stated before, our left hand side is a perfect square trinomial. But what is the binomial for it? It is the square root of the FIRST TERM plus the square root of the LAST TERM. The \(\sqrt{x^2} = x\) and the \(\sqrt{\frac{b^2}{4a^2}} = \frac{b}{2a}\). So, we have...

\[\Large (x + \frac{b}{2a})^2 = -\frac{4ac + b^2}{4a^2}\]
Now what? Now, we take the square root of both sides to isolate x.


\[\Large \sqrt{(x + \frac{b}{2a})^2} = \sqrt{\frac{-4ac + b^2}{4a^2}}\]
\[\Large (x + \frac{b}{2a}) = \frac{\pm \sqrt{-4ac + b^2}}{2a}\]
Note that square roots ALWAYS have \(\pm\) Then, we "transpose" \(\frac{b}{2a}\).

\[\Large x = \frac{\pm \sqrt{-4ac + b^2}}{2a} - \frac{b}{2a}\]
Since 2a is a common factor, we add...

\[\Large x = \frac{-b \pm \sqrt{-4ac + b^2}}{2a}\]
Notice anything? This is the quadratic formula, and we just derived it using completing the square! Need some demonstration on numbers? Let's use an easy one. 2x^2 + 3x - 4 = 0. I will not go over details, just refer to the tutorial above for the details on the steps.

\[\Large 2x^2 + 3x - 4 = 0\]
\[\Large \frac{2}{2} x^2 + \frac{3}{2} x - \frac{4}{2} = 0\]
\[\Large x^2 + \frac{3}{2} x = 2\]
\[\Large x^2 + \frac{3}{2} x + \frac{9}{16} = 2 + \frac{9}{16}\]
\[\Large (x + \frac{3}{4})^2 = \frac{41}{16}\]
\[\Large x + \frac{3}{4} = \frac{\pm \sqrt{41}}{4}\]
\[\Large x = \frac{\pm \sqrt{41}}{4} - \frac{3}{4}\]
\[\Large x = \frac{-3 \pm \sqrt{41}}{4}\]

This is our final answer.

Answer 2

Very clear and articulate. I am sure this will be helpful.

Answer 3

i certainly hope so thanks :)

Answer 4

Such a boss!

Answer 5

hmm i'll take that as a compliment ^_^ haha thanks

Answer 6

Um, using this formula might be easier:
\[x^2+bx = \left(x+\frac{b}{2}\right) ^{2} -\left(\frac{b}{2}\right) ^{2} \]
\[x^2-bx = (x-\frac{b}{2})^{2} -\left(\frac{b}{2}\right)^{2} \]

Answer 7

"easier" lol

Answer 8

yes, than going through all of the steps :P

Answer 9

for experienced users it is easier...my tutorials are for beginners :p why would i make a tutorial for experienced people abt completing the square hahaha =)))

Answer 10

i doubt that beginners would derive the formula :P
The formula i gave was for beginners.

Answer 11

riiiighhtt :p

Answer 12

Mimi gets a medal cause he ain't igbiw.

Answer 13

Nice tutorial :)