August 18th, 2017, 11:20 AM  #31  
Math Team Joined: Jul 2011 From: North America, 42nd parallel Posts: 3,372 Thanks: 233  Quote:
Oh yeah , don't forget about the derivation using the Tschirnhausen substitution I presented also.  
August 18th, 2017, 11:22 AM  #32 
Math Team Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 13,954 Thanks: 987 
thread closed!!

August 18th, 2017, 06:12 PM  #33 
Global Moderator Joined: Dec 2006 Posts: 20,274 Thanks: 1959 
I see there's no response to my initial question.

August 19th, 2017, 05:33 AM  #34 
Senior Member Joined: Aug 2017 From: United Kingdom Posts: 307 Thanks: 101 Math Focus: Number Theory, Algebraic Geometry  As it was pretty much the only interesting thing in the thread, it was bound to get ignored! Just to nitpick, surely you want the $\pm$ sign to be the opposite in each particular case? By which I mean $\frac{b + \sqrt{b^24ac}}{2a} = \frac{2c}{b  \sqrt{b^24ac}}$ and $\frac{b  \sqrt{b^24ac}}{2a} = \frac{2c}{b + \sqrt{b^24ac}}$ 
August 19th, 2017, 07:43 AM  #35  
Math Team Joined: Jul 2011 From: North America, 42nd parallel Posts: 3,372 Thanks: 233  I thought your initial question was directed at the OP Quote:
I would also like to note that if one uses $ax^2 = bx + c$ Then the above alternative formula becomes $\displaystyle \frac{b\pm\sqrt{b^2+4ac}}{2a}$ = $\displaystyle \frac{2c}{b\pm\sqrt{b^2+4ac}}$ And this may also be useful  
August 22nd, 2017, 03:13 PM  #36  
Senior Member Joined: Oct 2013 From: New York, USA Posts: 632 Thanks: 85  Quote:
 
August 22nd, 2017, 04:34 PM  #37  
Senior Member Joined: Aug 2017 From: United Kingdom Posts: 307 Thanks: 101 Math Focus: Number Theory, Algebraic Geometry  Quote:
$\displaystyle \frac{(3)\pm\sqrt{(3)^2+4 \times (1) \times (2)}}{2 \times (1)}= 1,2$ as expected.  
August 22nd, 2017, 04:47 PM  #38 
Global Moderator Joined: Dec 2006 Posts: 20,274 Thanks: 1959 
What makes you think that, EvanJ? I think, agentredlum, that you didn't understand what can make the formulas I suggested particularly useful. Give it some deeper thought. 
August 24th, 2017, 12:01 AM  #39  
Math Team Joined: Jul 2011 From: North America, 42nd parallel Posts: 3,372 Thanks: 233  Quote:
But I'm on the fence about that. Maybe you can convince me?  
August 24th, 2017, 12:38 AM  #40 
Global Moderator Joined: Dec 2006 Posts: 20,274 Thanks: 1959 
Try using an extreme example to test that.


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