Can someone help explain 'completing the square' to me? I'm so lost

Question

whpalmer4 · Answer

Yes. 

When we complete the square, we are regrouping part of our equation or expression so that it can be written as the square of a binomial, such as $(x+a)^2$

whpalmer4 · Answer

Let's do some homework to see how we will accomplish this sleight of hand.

$$(x+a)^2=(x+a)(x+a)=x^2+a*x+a*x+a*a$$$$\qquad=x^2+2ax+a^2$$

whpalmer4 · Answer

So what this means for us is that when we have something like $$x^2+6x$$we can find the constant we need to add (or steal from some other expression) to complete the square by taking half of the coefficient of the $x$ term.  If we call that number $a$, we can write our completed square as $(x+a)^2$, and we need to either add $a^2$ to both sides of the equation, or subtract $a^2$ from some quantity on the same side of the = sign.  I'll do an example each way.

whpalmer4 · Answer

For the example I mentioned, $x^2+6x$, the coefficient of $x$ is 6. Half of 6 is 3, so $a=3$, and we need to combine the two terms we already have, $x^2+6x$, with \a^2=3^2=9:

$$x^2+6x+9=(x+3)^2$$

Let's check by multiplying:$$(x+3)^2=(x+3)(x+3)=x(x+3)+3(x+3)=x^2+3x+3x+3*3$$$$\qquad=x^2+6x+9$$

whpalmer4 · Answer

Now let's do this in the context of an equation.

$$x^2+2x=0$$here our coefficient of $x$ is 2, so $a=2/1=1$ and we will add $a^2=1^2=1$ to each side of the equation.
$$x^2+2x+1=0+1$$we can rewrite the left side now:$$(x+a)^2=(x+1)^2=1$$

If we were interested in finding the values of $x$, at this point we would take the square root of each side and solve the resulting equation for the value(s) of $x$.

anonymous · Answer

Thank You sooooo Much!!!!

whpalmer4 · Answer

Sorry, I wrote an additional section, but lost it to the OpenStudy gremlins this morning.  

Here's the other case, where you extract that number needed to complete the square from something that is already present.

$$x^2+10x +9 =0$$Half of the coefficient of $x$ ($10$) is $5$, and squared that makes $25$.  Our completed square will be $(x+5)^2$ but we need to get that $25$ from somewhere.  We'll take it from the $9$ which is just sitting there, defenseless :-)

$$(x^2+10x + 25) +9 - 25 = 0$$Note that I've added 25 within the parentheses (which will turn into our completed square in the next step), and subtracted it outside the parentheses.  $25-25=0$, of course, so I haven't changed the value of anything by doing that, but it makes it clear what is going on.

$$(x+5)^2 + 9-25 = 0$$$$(x+5)^2-16 = 0$$and if we were solving for $x$ our next steps would be $$(x+5)^2 = 16$$$$(x+5) = \pm\sqrt{16}$$$$x+5 = \pm 4$$$$x = -1,~x=-9$$

My apologies if the part about solving for the values of $x$ goes beyond where you've gotten so far, leading to potential confusion.  You'll see it soon enough :-)

Completing the square is also how the "quadratic formula" is derived.  Left as an exercise for the reader, as they say in the textbook writing biz :-)

anonymous · Answer

Haha Thanks! Yah, it's still clear as dirt, but this is helping me understand

whpalmer4 · Answer

Think of it as trying to make a square array of blocks out of a collection of blocks on the floor.  You may have enough blocks to make a square, in which case, you just take the number you need and leave the rest.  Or, you may need a few more, in which case, your mom comes along and gives both you and your playmate some extra blocks, so you both have the same number of blocks as each other, and you have enough to make your square :-)