## Completing the square

Are you an expert at completing the square? This is an important skill that is not nearly as difficult as a lot of students think it is. If you master just a couple of rules, you will be able to complete squares like a pro.

Let’s start with an explanation. Since we complete the square in algebra and not in geometry, exactly where is the square we are completing? This confuses many students and is probably the reason why they are frightened unnecessarily by the process. But it’s actually pretty simple. Let’s say we have the following expression:

$(x-3)^2$

This is obviously the square of $(x-3)$, and so we call it a perfect square. If we expand the expression into $x^2-6x+9$, we recognize that it is still a perfect square. Any trinomial of the form $ax^2+bx+c$ is a perfect square if it is the square of a binomial expression.

When we have an expression such as $x^2-6x$, we look for a number we can add to make it a perfect square of a binomial. In this case, adding $9$ to $x^2-6x$ to get $x^2-6x+9$ gives us the square of $x-3$. Therefore we have completed the square. That’s all the expression means!

Completing the square is actually quite easy in most cases. Because $(x+a)^2=x^2+2ax+a^2$, anytime we have an expression of the form $x^2+2ax$, all we need to do is add $a^2$. You can do this in two quick steps:

1. Take the coefficient on the $x$ term and divide it by $2$.
2. Square this number and add it to the end.

For example:

• $x^2+8x$: Divide the $8$ by $2$ (you get $4$) and square it (you get $16$), so the completed square is $x^2+8x \; +$ $16$
• $x^2-12x$: Divide the $-12$ by $2$ (you get $-6$) and square it (you get $36$), so the completed square is $x^2-12x \; +$ $36$
• $x^2+9x$: Divide the $9$ by $2$ (you get $\frac{9}{2}$) and square it (you get $\frac{81}{4}$), so the completed square is $x^2+9x \; +$ $\frac{81}{4}$
• $x^2-13x$: Divide the $-13$ by $2$ (you get $\frac{-13}{2}$) and square it (you get $\frac{169}{4}$), so the completed square is $x^2-13x \; +$ $\frac{169}{4}$

Next you write the factor as a square. The factor is always of the form $(x+a)^2$ or $(x-a)^2$ where the value of $a$ is the number that you squared. For the four examples above:

• $x^2+8x \; +$ $16$ $= (x$ $+ \; 4$ $)^2$
• $x^2-12x \; +$ $36$ $= (x$ $- \; 6$ $)^2$
• $x^2+9x \; +$ $\frac{81}{4}$ $= (x$ $+ \; \frac{9}{2}$ $)^2$
• $x^2-13x \; +$ $\frac{169}{4}$ $= (x$ $- \; \frac{13}{2}$ $)^2$

What if your expression starts with $ax^2$ where $a$ is not equal to $1$? Then you have to factor the $a$ out of both terms before you complete the steps above. It can appear very complicated, but the process is the same every time. Here are some examples:

• $2x^2+8x=2(x^2+4x) \rightarrow 2(x^2+4x+4)=2(x+2)^2$
• $4x^2-12x=4(x^2-3x) \rightarrow 4(x^2-3x+ \frac{9}{4}) = 4(x- \frac{3}{2})^2$
• $2x^2+9x=2(x^2+ \frac{9}{2}x) \rightarrow 2(x^2+ \frac{9}{2}x+ \frac{81}{16}) = 2(x+ \frac{9}{4})^2$
• $3x^2-13x=3(x^2- \frac{13}{3}x) \rightarrow 3(x^2- \frac{13}{3}x+ \frac{169}{36})=3(x- \frac{13}{6})^2$
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