Algebra 1 Substitution
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Equations in Disguise

 So now we have a great way to solve equations that look like:

 2X2 + 4X - 6 = 0

 What about stuff like ...

 If you look really close,
 you will see that the exponents on the X terms on the left
 are exactly the squares of the exponents on the X terms in the center.
 We know that ...

 X2 = X X

 But these are also squares ...

 Big deal. So what?
 OK, here's what we do.
 We chose a letter (usually "u")
 to stand for the X term in the middle of the equation.
 That way, the X term on the left is u2.
 Here goes ...
 Solve this puppy ...

 Now substitute "u" and "u 2" into the equation for the X's ...

u2 - 6u + 5 = 0

 The Quadratic Formula works for any letter, not just X.
 That means we can use it here.
 Lucky us!


 But u = , so now put the back in for u and find X ...

 Let's do another one ...

 u = X 2 so we put X 2 back in to find X ...

 Since there are not any real numbers that you can square 
 to get a negative three, that one has no answer.
 So the only answer to this one is X = 1
 These things could get really involved.
 The trick is to just look at the thing you replace with the u
 as a glob.
 Then just forget about it until you get done
 with the rest of the problem.
 Here's a nasty one.

4(2X2 + 6X)2 + 12(2X2 + 6X) + 5 = 0

 Substitute ...

 u = (2X2 + 6)   so   u2 = (2X2 + 6)2

 We have ...

 So great, we found u.
 Now we need to put back the thing that we substituted for
 (2X 2 + 6X) so ...

 Now move the numbers over to the left side of the equation
 so that we can use the Quadratic Formula ...

 So ...

 That one was pretty messy.
 But the deal is that no matter what kind of mess you have,
 if the variable glob on the left is the square 
 of the variable glob in the middle you can use this trick.

   copyright 2005 Bruce Kirkpatrick

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