Algebra 1 Factoring by Completing the Square
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Complete That Square
Factoring by Completing the Square

 
 Once upon a time a few pages ago,
 I promised you a method to factor some equations directly
 without all of the guessing games.
 
 Here it is.
 
 Say you have an equation that looks something like ...
 

 6X2 + 2X - 4 = 0

 
 We will be trying to turn this into ...
 

(X some number)2 = some other number

 
 So how do we do it?
 
 Well, when we had (X + 4)2 and multiplied it out,
 we got ...

(X + 4)2 = X2 + 8X + 16

 
 In general, using the letter "B" to stand for 
 any old number that we might have
 we get ...

(X + B)2 = X2 + 2BX + B2

 
 This is what we need to get to.
 We take whatever we have at the start
 and make it look like X 2 + 2BX + B 2
 
 So looking at 6X 2 + 2X - 4 = 0, 
 the first thing we need to do is get rid of the 6 part of the 6X 2.

To do that, we divide both sides by 6 ...

 
 STEP 1:
 Divide both sides of the equation by the coefficient on the X 2 ...

 

 
 and simplify ...
 

 
 STEP 2:
 Take the X term and determine the value of B.
 Use that to find B 2.
 
 In this problem, the X term is 1/3X. 
 Our formula says this is equal to 2BX.
 Solve for B ...
 

 
 So if 1/6 is B then multiply it times itself
 to get B 2.

 
 STEP 3:
 Put the B 2 value into the equation.
 
 We can do almost anything we want to an equation
 if we do the same thing to both sides.
 

 
 STEP 4:
 Gather up the terms of the square, 
 and write them as the square.
 
 At this point, the first three terms on the left side
 are the terms of our square.
 That is, the (X + B) 2 = X 2 + 2BX + B 2
 We can substitute the square for these three terms ...
 

 
 STEP 5:
 Put all of the other baloney that's not part of the square 
 on the right side of the equation.
 Simplify it as much as you can.
 

 
 STEP 6:
 Solve this for X.
 
 We're finally at the place where we have ...
 

(X some number)2 = some other number

 
 From here it's easy.
 We just peel the onion ...
 

 
 So we have two answers ...
 

 
 If you check back a few pages,
 you'll see that we got the same answer there too!
 

   copyright 2005 Bruce Kirkpatrick

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