August 16th, 2017, 07:43 AM  #11  
Math Team Joined: Jul 2011 From: North America, 42nd parallel Posts: 3,372 Thanks: 233  Quote:
$ \ \ \ \ \ $ That's incorrect cjem Quote:
Nowhere in my proof did I do what you claim I did Verbatim? Do you know meaning of the word verbatim? I completed the square on an abstract 2nd degree equation , that is NOT the same as relabeling coefficients Did you read the proof? Are you able to follow the proof? Can you find any mistakes in the proof? Do you have any questions about the derivation of the proof? Quote:
I'm pretty sure you are not the authority on what is or is not interesting in mathematics cjem. Last edited by skipjack; August 16th, 2017 at 09:40 PM.  
August 16th, 2017, 07:59 AM  #12 
Senior Member Joined: Aug 2017 From: United Kingdom Posts: 311 Thanks: 109 Math Focus: Number Theory, Algebraic Geometry  You started with the equation $ax^2 = bx + c$ (which is equivalent to $ax^2  bx  c = 0$) rather than $ax^2 + bx + c = 0$, which is what is normally used. That is all I meant by "relabeling coefficients"  you're using $b$ and $c$, where $b$ and $c$ are normally used, respectively. Of course your proof is correct and easy to follow  it's the same as the usual one, barring that one difference. Well, technically you haven't actually written a proof  you've just written a list of equations without indicating any logical implications (but that could be fixed simply by adding a "$\implies$" or the word "so" at the start of every line except the first, so I won't dwell on this!) Last edited by cjem; August 16th, 2017 at 08:10 AM. 
August 16th, 2017, 08:28 AM  #13 
Math Team Joined: Jul 2011 From: North America, 42nd parallel Posts: 3,372 Thanks: 233 
cjem , where are you from? Half an hour ago your profile said Paris now it says United Kingdom. Yes , the proof is very easy to follow but I don't think you can follow it because you keep insisting on making the same incorrect argument. I did NOT set the equation to zero , I did NOT rename coefficients. If I gave you a proof using a Tschirnhausen Transformation would you still claim I am renaming coefficients? 
August 16th, 2017, 08:36 AM  #14  
Math Team Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 14,112 Thanks: 1002  Quote:
Isn't one formula enough?  
August 16th, 2017, 08:52 AM  #15  
Senior Member Joined: Aug 2017 From: United Kingdom Posts: 311 Thanks: 109 Math Focus: Number Theory, Algebraic Geometry  $ax^2 = bx + c \iff ax^2  bx  c = 0$. That is, the statement $ax^2 = bx + c$ is logically equivalent to the statement $ax^2  bx  c = 0$. I'm probably just wording it badly. The normal equation used is $ax^2 + bx + c = 0$, which is equivalent (by equivalent, I mean logically equivalent) to $ax^2 = bx  c$. You've used the equation $ax^2 = bx + c$. Notice in your equation that the signs of the linear and constant terms are reversed compared to the normal one? That's all I mean by "relabeling coefficients". Quote:
The thing is, what you've actually done is rearranged your equation to get the discriminant one side and completed the square in the other side. Then you've taken the squareroot of both sides (taking positive and negative roots into account), and further rearranged to find an expression for $x$. This is exactly the same argument that is usually used. I'm not sure what relevance my location has to this discussion, but I'm from the UK.  
August 16th, 2017, 11:03 AM  #16 
Math Team Joined: Jul 2011 From: North America, 42nd parallel Posts: 3,372 Thanks: 233 
cjem , 'equivalent' and 'exactly' do not mean the same thing. Just because you see the destination and you have an idea about how to get there does not prove I took the route you think I did. You are still trying to defend your original incorrect assumptions about what I did. Anyway , here's a derivation using a Tschirnhausen substitution. For all $ \ \ \ a \ne 0 $ If $ \ \ \ ax^2 = bx + c $ Then $ \ \ \ x = \frac{b \pm \sqrt{b^2 + 4ac}}{2a} $ Proof by using a Tschirnhausen Transformation $ ax^2 = bx + c \ \ \ \ $ , Let $ \ \ \ \ x = y + \frac{b}{a} $ $ a(y + \frac{b }{a})^2 = b( y + \frac{b }{a}) + c $ $(y + \frac{b}{a})^2 = \frac{by}{a} + \frac{b^2}{a^2} + \frac{c}{a} $ $ y^2 + \frac{2by}{a} + \frac{b^2}{a^2} = \frac{by}{a} + \frac{b^2}{a^2} + \frac{c}{a}$ $y^2 + \frac{by}{a} = \frac{c}{a} $ $y^2 + \frac{by}{a} + \frac{b^2}{4a^2} = \frac{b^2}{4a^2} + \frac{c}{a} $ $ (y + \frac{b}{2a} )^2 = \frac{b^2 + 4ac}{4a^2} $ $ y + \frac{b}{2a} = \pm \frac{ \sqrt{b^2 + 4ac}}{2a} $ But $ \ \ \ x = y + \frac{b}{a} \ \ \ $ , $ \ \ \ y = x  \frac{b}{a} $ So ... $ x  \frac{b}{a} + \frac{b}{2a} = \pm \frac{\sqrt{b^2 + 4ac}}{2a} $ $x  \frac{b}{2a} = \pm \frac{\sqrt{b^2 + 4ac}}{2a} $ $ x = \frac{b \pm \sqrt{b^2 + 4ac}}{2a} $ Q.E.D. 
August 16th, 2017, 11:29 AM  #17  
Senior Member Joined: Aug 2017 From: United Kingdom Posts: 311 Thanks: 109 Math Focus: Number Theory, Algebraic Geometry  Quote:
Quote:
Quote:
Nice, I'm sure this argument will be of interest to people here. In fact, the first derivation of the cubic formula I saw used a Tschirnhausen transformation  it makes it quite straightforward to do, even with just elementary algebra.  
August 16th, 2017, 11:59 AM  #18 
Math Team Joined: Jul 2011 From: North America, 42nd parallel Posts: 3,372 Thanks: 233 
Of course you are free to post your own interpretation and you certainly don't need my permission to post your own interpretation but you made incorrect claims about what I did in post #5. What I did is shown in the presentation of post #5. I did not do the 'things' you claim I did in post #5. I think I've already made my arguments about why your assumptions are incorrect , I don't want to rehash. It's all good , let's leave it alone. 
August 16th, 2017, 12:07 PM  #19  
Senior Member Joined: Aug 2017 From: United Kingdom Posts: 311 Thanks: 109 Math Focus: Number Theory, Algebraic Geometry  Quote:
 
August 16th, 2017, 09:32 PM  #20 
Global Moderator Joined: Dec 2006 Posts: 20,370 Thanks: 2007  Code: agentredlum Usual derivation axÂ² = bx + c axÂ² + bx + c = 0 axÂ²  bx = c axÂ² + bx = c 4aÂ²xÂ²  4abx = 4ac 4aÂ²xÂ² + 4abx = 4ac 4aÂ²xÂ²  4abx + bÂ² = bÂ² + 4ac 4aÂ²xÂ² + 4abx + bÂ² = bÂ²  4ac (2ax  b)Â² = bÂ² + 4ac (2ax + b)Â² = bÂ²  4ac 2ax  b = Â±âˆš(bÂ² + 4ac) 2ax + b = Â±âˆš(bÂ²  4ac) x = (b Â±âˆš(bÂ² + 4ac))/(2a) x = (b Â±âˆš(bÂ²  4ac))/(2a) The differences are a slightly different first equation and a few sign changes. Why is the result interesting? 

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