Questions about quadratic formula

AI Thread Summary
The discussion revolves around the quadratic formula and the occurrence of double and triple negatives in its computation. Participants debate the probability of encountering double negatives for -b and triple negatives for -4ac when given random second-degree equations. It is suggested that if coefficients a, b, and c have equal probabilities for being positive or negative, the expected occurrence of double negatives is 50%, while triple negatives would occur 25% of the time. The conversation emphasizes the need for a defined probability distribution when discussing random equations, as the infinite nature of real numbers complicates the idea of percentages. Ultimately, the focus remains on understanding the implications of sign combinations in quadratic equations.
  • #51
Robert1986 said:
I never ever said I came up with the "trick"; I gave a very simple proof that your trick worked. There is nothing at all dishonest about that. The proof is, in fact, mine.

Who gets credit for the proof of FLT? Fermat or Wiles?

I rest my case.

To be honest you can't mention Wiles without mentioning Fermat

You used the 'trick' i posted without mentioning where you got it.
 
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  • #52
Robert1986 said:
Well, it has been my experience (again, this is just my experience) that after using the QF for about 3 weeks, it is no longer necessary for most students to explicitly write out what they are doing.

Well, if you work at a learning center where every day you have to show students STEP BY STEP how to use a formula... you might start thinking about some shortcuts.

My own personal opinion of the matter is...If the people writing the texts knew it could be done that way... they would show it.

It is folly to do with more when you could do with less- Occam's Razor
 
  • #53
agentredlum said:
To be honest you can't mention Wiles without mentioning Fermat
That wasn't my question. My question was who gets credit for the proof. It was Fermat's idea but Wiles gave a proof. The "trick" is your idea, but I gave a proof. Therefore, I get credit (but not much credit because it is, after all, a trivial thing to prove) for the proof I gave. It is pretty simple.


agentredlum said:
You used the 'trick' i posted without mentioning where you got it.

I never use your trick. And it is pretty obvious where I got it, man. I never claimed that I came up with the trick; I only claimed that I gave a proof of it. Do you dispute this? Do you need me to go to the thread and post my proof?

I think it is you who has a difficult time understanding the English language.
 
  • #54
agentredlum said:
If the people writing the texts knew it could be done that way... they would show it.

So you think you are smarter than the people who write the math texts?? Ok...

Anyway, the reason that the original quadratic formula is used because of the definition of a root.
 
  • #55
agentredlum said:
Well, if you work at a learning center where every day you have to show students STEP BY STEP how to use a formula... you might start thinking about some shortcuts.

My own personal opinion of the matter is...If the people writing the texts knew it could be done that way... they would show it.

Yeah, James Stewart missed your little trick; that's why he didn't mention it.


agentredlum said:
It is folly to do with more when you could do with less- Occam's Razor

Exactly. This is why I prefer to remember one QF and not 4.
 
  • #56
micromass said:
What I cannot understand is this: how can somebody who completed calc I,II,III and linear algebra think that your trick is nontrivial and useful?? I really cannot grasp that.
Everybody I know would call your trick a trivial result and wouldn't defend it as hard as you do.

It's like saying that the quadratic equation

ax^2+bx+c=0

has a root given by

\frac{\sqrt{b^2-4ac}-b}{2a}

and somehow claim that above formula does not follow trivially from the quadratic formula and that the formula is useful somehow...

You are using the word trivial to mean useless.

They don't mean the same thing.

The result is trivial in the sense that you get it quickly without higher math involved.

Useless is a matter of opinion.
 
  • #57
agentredlum said:
You are using the word trivial to mean useless.

No, I didn't.
 
  • #58
Robert1986 said:
This is why I prefer to remember one QF and not 4.

You keep saying this, even though I keep correcting you.

I don't use 4 formulas.

I only use 1 formula, the formula that minimizes operations.

You refuse to understand this-Occam's Razor
 
  • #59
micromass said:
No, I didn't.

Many times, in many posts you and others said there was no use for it. Even only as a curiosity it is useful. It has computational advantages as well.

So what are you saying now?

It is trivial but useful?
 
  • #60
agentredlum said:
Many times, in many posts you and others said there was no use for it. Even only as a curiosity it is useful. It has computational advantages as well.

So what are you saying now?

It is trivial but useful?

No, I'm saying it's trivial AND useless. I'm not mixing up the two words.
 
  • #61
agentredlum said:
You keep saying this, even though I keep correcting you.

I don't use 4 formulas.

I only use 1 formula, the formula that minimizes operations.

You refuse to understand this-Occam's Razor

Ahh, I see. But it only minimizes operations when you have a quadratic in one particular form, correct?


Anyway, I prefer the formula that is derived directly using the definition of a root of a function.
 
  • #62
agentredlum said:
Many times, in many posts you and others said there was no use for it. Even only as a curiosity it is useful. It has computational advantages as well.

So what are you saying now?

It is trivial but useful?

I can't speak for micromass or anyone else, but my claim is that it is trivial and usless.
 
  • #63
Robert1986 said:
Ahh, I see. But it only minimizes operations when you have a quadratic in one particular form, correct?Anyway, I prefer the formula that is derived directly using the definition of a root of a function.

I NEVER have to compute a double negative for b

I NEVER have to compute a triple negative for +4ac

5 out of 8 times you have problems i don't

Here is a fact about amicable numbers.

In 1636, Fermat found the pair (17296, 18416) and in 1638, Descartes found (9363584, 9437056), although these results were actually rediscoveries of numbers known to Arab mathematicians. By 1747, Euler had found 30 pairs, a number which he later extended to 60. In 1866, 16-year old B.*Nicolò I.*Paganini found the small amicable pair (1184, 1210) which had eluded his more illustrious predecessors (Paganini 1866-1867; Dickson 2005, p.*47).

http://mathworld.wolfram.com/AmicablePair.html

This 16 year old boy found something that escaped all the geniuses that came before him.
 
  • #64
agentredlum said:
5 out of 8 times you have problems i don't

The point is that we don't see minus signs as problems. A minus less or more means nothing to us...
 
  • #65
agentredlum said:
I NEVER have to compute a double negative for b

I NEVER have to compute a triple negative for +4ac

5 out of 8 times you have problems i don't

But this is my point, doing a double negative isn't, in any way, a problem for me. "Computing" a double negative is as simple as NOT writting a "-". It is easy.


agentredlum said:
Here is a fact about amicable numbers.

In 1636, Fermat found the pair (17296, 18416) and in 1638, Descartes found (9363584, 9437056), although these results were actually rediscoveries of numbers known to Arab mathematicians. By 1747, Euler had found 30 pairs, a number which he later extended to 60. In 1866, 16-year old B.*Nicolò I.*Paganini found the small amicable pair (1184, 1210) which had eluded his more illustrious predecessors (Paganini 1866-1867; Dickson 2005, p.*47).

http://mathworld.wolfram.com/AmicablePair.html

This 16 year old boy found something that escaped all the geniuses that came before him.


This is not what is going on, here.
 
  • #66
Robert1986 said:
Ahh, I see. But it only minimizes operations when you have a quadratic in one particular form, correct?

It minimizes operations in 5 out of 8 forms

62.5% of the time, i do less operations.

THAT IS A PHENOMENAL RESULT!
 
  • #67
micromass said:
The point is that we don't see minus signs as problems. A minus less or more means nothing to us...

What does it mean to a computer, in terms of memory allocation?
 
  • #68
agentredlum said:
What does it mean to a computer, in terms of memory allocation?

Aah, I was hoping that you would come to this! :biggrin:

The thing is that a computer keeps a bit that signifies the sign of a number. 0 if it's positive, 1 if it's negative. So whether a number is positive or negative: there is a bit allocated to the sign of the number! So your formula makes no difference at all in terms of computer memory.

And even if it did: I don't think a 30GB computer would have trouble with a little minus sign...
 
  • #69
agentredlum said:
What does it mean to a computer, in terms of memory allocation?

Haha. If you are doing standard arithmetic, it means nothing. The highest bit is the sign bit. 1 for negative and 0 for positive.
 
  • #70
This thread becomes trivial and useless. Locked.
 

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