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Bronze Enthusiast
Picture of silenteuphony
Posted
As a teacher/tutor, I've picked up a few mathematical shortcuts over the years, including one for factoring quadratic equations. I wanted to share my method, and see if anyone else knows any interesting shortcuts.

Given a quadratic equation with a leading coefficient other than 1, such as

Ax² + Bx + C  =  0

you can use the transformation  u = Ax  to "shift" the A to the last term and get the simpler equation

u² + Bu + ACu  =  0

Then you factor this simpler equation to solve for u, and divide by A to get  x = u/A



This method is excellent for equations with fractional solutions. Here's a simple example with actual numbers:

6x² + x - 2  =  0

u² + u - 12  =  0

(u + 4) (u - 3)  =  0

u = -4  or  u = 3

x = -4/6  or  x = 3/6

x = -2/3  or  x = 1/2


Of course, many beginning algebra students are uncomfortable with the change of variable; it's okay to keep the same variable as long as you remember to divide your solutions by the leading coefficient at the end of the problem.

I've seen a couple methods that are based on the same type of logic, but have different approaches. Does anybody else have a favorite method that might help struggling algebra students?
 
Posts: 265 | Location: Denver, Colorado, USA | Registered: 06-04-02Reply With QuoteEdit or Delete MessageReport This Post
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Wait a sec... u² + Bu + ACu = 0 when factored with Ax = A²x² + ABx + A²Cx = 0 ... I understand this is multiplication by A, but that last term doesn't seem to fit: the A²Cx...

EDIT: Oops, nevermind... I see what it was, but shouldn't the equation read: u² + Bu + Ac = 0?
 
Posts: 1363 | Location: Lowell, MA, USA | Registered: 06-03-02Reply With QuoteEdit or Delete MessageReport This Post
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Picture of silenteuphony
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Thanks for the catch, WK, I did stick an extra u in. The transformed equation should read:

u² + Bu + AC  =  0

Incidentally, this method will work with any quadratic equation with rational solutions. That is, given any quadratic equation with integer coefficients and rational solutions, u will always have integer solutions.
 
Posts: 265 | Location: Denver, Colorado, USA | Registered: 06-04-02Reply With QuoteEdit or Delete MessageReport This Post
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One way I learned to factor was to take all positive and negative factors of C and put each one over each positive or negative factor of A. These are all possible rational solutions. For instance, in 3x²+4x-4, the only possible rational factors are 1, -1, 2, -2, 4, -4, 1/3, -1/3, 2/3, -2/3, 4/3, or -4/3. We wind up with (3x-2)(x+2) or x= -2/3, 2.

Usually we'd write it

+/- 1, 2, 4
------------
+/- 1, 3
 
Posts: 1363 | Location: Lowell, MA, USA | Registered: 06-03-02Reply With QuoteEdit or Delete MessageReport This Post
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