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August 16th, 2017, 07:43 AM   #11
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Quote:
 Originally Posted by cjem ...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's incorrect cjem

Quote:
 Originally Posted by cjem 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.
You statements are incorrect.

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

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:
 Originally Posted by cjem 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...
$\ \ \ \ \$ Your statement is incorrect and based on your opinion. I'm pretty sure that no one is an authority on what is or is not interesting in mathematics. I posted a derivation which I thought the OP may find interesting. That's all ... I'm not trying to 'pass off' anything as anything.

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
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Quote:
 Originally Posted by agentredlum That's incorrect cjem
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
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Quote:
 Originally Posted by Antoniomathgini 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 $\displaystyle \frac{b\pm\sqrt{b^2-4ac}}{-2a}$. Is this an alternative quadratic formula?
Suppose it is; so what?
Isn't one formula enough?

August 16th, 2017, 08:52 AM   #15
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Quote:
 Originally Posted by agentredlum I did NOT set the equation to zero
$ax^2 = bx + c \iff ax^2 - bx - c = 0$. That is, the statement $ax^2 = bx + c$ is logically equivalent to the statement $ax^2 - bx - c = 0$.

Quote:
 Originally Posted by agentredlum I did NOT rename coefficients.
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:
 Originally Posted by agentredlum If I gave you a proof using a Tschirnhausen Transformation would you still claim I am renaming coefficients?
I would not.

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.

Quote:
 Originally Posted by agentredlum cjem , where are you from? Half an hour ago your profile said Paris now it says United Kingdom.
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
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Quote:
 Originally Posted by agentredlum cjem , 'equivalent' and 'exactly' do not mean the same thing.
Correct.

Quote:
 Originally Posted by agentredlum 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.
I read your argument and summarized the key steps you used. If you feel I've done this inaccurately, I'd be happy for you to correct me.

Quote:
 Originally Posted by agentredlum You are still trying to defend your original incorrect assumptions about what I did.
Would you mind explaining what assumptions I've made and why they are incorrect?

Quote:
 Originally Posted by agentredlum 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 ...
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
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Quote:
 Originally Posted by agentredlum 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.
I'm still not even sure what assumptions you think I've made. I'd be happy to continue this discussion in PMs to try and straighten it out, but I'd be equally happy to just drop it :P

 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) In both cases, the steps are (1) rearrange, (2) multiply by 4a, (3) add bÂ² to each side, (4) write the LHS as a square, (5) take the square root of each side, and (6) rearrange to isolate x. The differences are a slightly different first equation and a few sign changes. Why is the result interesting? Thanks from Denis

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