August 15th, 2017, 06:42 AM  #1 
Member Joined: Apr 2017 From: Canada Posts: 32 Thanks: 2  Alternative Quadratic Formula
I stumbled upon something interesting. I don't think it's anything new, but I've never been shown this. When deriving the quadratic formula there is a part where it is necessary to take the square root of $\displaystyle 4a^2$. Previously I've always automatically written $\displaystyle 2a$. However, I now realize that it could just as well have been $\displaystyle 2a$. When we do write $\displaystyle 2a $, it changes our final formula to $\displaystyle \frac{b\pm\sqrt{b^24ac}}{2a}$. Is this an alternative quadratic formula? Last edited by skipjack; August 15th, 2017 at 07:39 AM. 
August 15th, 2017, 07:15 AM  #2 
Senior Member Joined: Aug 2017 From: United Kingdom Posts: 270 Thanks: 81 Math Focus: Algebraic Number Theory, Arithmetic Geometry 
That's exactly the same as the one you're used to.

August 15th, 2017, 07:53 AM  #3 
Global Moderator Joined: Dec 2006 Posts: 19,725 Thanks: 1807 
It can be useful to use both $\displaystyle \frac{b\pm\sqrt{b^24ac}}{2a}$ and $\displaystyle \frac{2c}{b\pm\sqrt{b^24ac}}$, choosing the Â± sign in the same particular way in each case. Can you see why?

August 15th, 2017, 06:22 PM  #4  
Senior Member Joined: May 2016 From: USA Posts: 1,148 Thanks: 479  Quote:
$x^2 + 2 * \dfrac{b}{2a} + \left ( \dfrac{b}{2a} \right )^2 = \left ( \dfrac{b}{2a} \right )^2  \dfrac{4ac}{4a^2} \implies$ $\left ( x + \dfrac{b}{2a} \right )^2 = \dfrac {b^2  4ac}{4a^2} \implies x + \dfrac{b}{2a} = \pm \sqrt{\dfrac{b^2  4ac}{4a^2}} = \dfrac{\pm \sqrt{b^2  4ac}}{2a} \implies$ $x = \dfrac{\ b \pm \sqrt{b^2  4ac}}{2a} \ne \dfrac{b \pm\sqrt{b^2  4ac}}{\ 2a}$ $\text {UNLESS } b = 0 = c.$ What you mean is that $x = \dfrac{\ b \pm \sqrt{b^2  4ac}}{2a} = \dfrac{\ b \pm \sqrt{b^2  4ac}}{2a} * \dfrac{\ 1}{\ 1} = \dfrac{b \mp \sqrt{b^2  4ac}}{\ 2a}.$ There are not two different formulas. Rather there are two different ways of expressing the same formula. I am being persnickety, but math's strength comes from being precise.  
August 16th, 2017, 01:00 AM  #5 
Math Team Joined: Jul 2011 From: North America, 42nd parallel Posts: 3,372 Thanks: 233 
Hi , your formula works and does not look the same as the standard formula so yes it is an alternative version. Now , regarding 'things' you've never been shown , perhaps you may find the following interesting. For all $ \ \ \ a \ne 0$ If $ \ \ \ \ \ ax^2 = bx + c$ Then $ \ \ \ \ \ x = \frac{b \pm \sqrt{b^2 + 4ac}}{2a} $ Proof : $ax^2 = bx + c$ $ax^2  bx = c$ $4a^2x^2  4abx = 4ac$ $4a^2x^2 4abx + b^2 = b^2 + 4ac$ $ (2ax  b)^2 = b^2 + 4ac$ $ 2ax  b = \pm \sqrt{b^2 + 4ac} $ $ 2ax = b \pm \sqrt{b^2 + 4ac} $ $ x = \frac{b \pm \sqrt{b^2 + 4ac} }{2a} $ Q.E.D. Last edited by agentredlum; August 16th, 2017 at 01:07 AM. 
August 16th, 2017, 04:46 AM  #6 
Senior Member Joined: Aug 2017 From: United Kingdom Posts: 270 Thanks: 81 Math Focus: Algebraic Number Theory, Arithmetic Geometry  The "normal" formula gives the solutions to the equation $ax^2 + bx + c = 0$. All you've done is relabeled two of the coefficients (you've called the linear coefficient $b$ rather than $b$ and the constant one $c$ rather than $c$) and done the standard proof. That does not make it different in any significant way.

August 16th, 2017, 05:22 AM  #7  
Senior Member Joined: Aug 2017 From: United Kingdom Posts: 270 Thanks: 81 Math Focus: Algebraic Number Theory, Arithmetic Geometry  Quote:
 
August 16th, 2017, 06:00 AM  #8 
Math Team Joined: Jul 2011 From: North America, 42nd parallel Posts: 3,372 Thanks: 233 
The proof stands alone. "Different in any significant way" Is just your opinion 
August 16th, 2017, 06:20 AM  #9  
Senior Member Joined: May 2016 From: USA Posts: 1,148 Thanks: 479  Quote:
Fine. Consider the case where the discriminant is zero. Using the normal formula we get $x = \dfrac{\ b}{2a}.$ That has the exact same meaning as $x = \dfrac{b}{\ 2a}.$ To view them as having some sort of mysterious difference is not recognizing the ambiguity in plus/minus but being engulfed in it. The distinction is obviously trivial.  
August 16th, 2017, 06:55 AM  #10 
Senior Member Joined: Aug 2017 From: United Kingdom Posts: 270 Thanks: 81 Math Focus: Algebraic Number Theory, Arithmetic Geometry  You have copied the usual proof verbatim  you've made exactly the same logical steps in exactly the same order. The only difference is that you've decided to change the symbols for two of the coefficients. To pass this off as something different or interesting is to miss the point  in mathematics, it doesn't matter what symbols we use for things...
Last edited by cjem; August 16th, 2017 at 07:03 AM. 

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