Questions about quadratic formula

[agentredlum]

Proof for that, please?

Answer the question

 

[micromass]

Answer the question

You didn't ask any question.
You made an assertion, which I ask you to prove.

 

[agentredlum]

You didn't ask any question.
You made an assertion, which I ask you to prove.

Here it is AGAIN

What is the probability that you have something that factors nicely?

 

[micromass]

Here it is AGAIN

What is the probability that you have something that factors nicely?

That probability is 1. I can factor each quadratic equation.

Your turn: prove your previous assertion.

 

[Robert1986]

Assuming uniform probability distribution for the co-efficients.

What is the probability that you have something that factors nicely?

If it is more than 0 + delta(e) where delta(e) goes to zero, I would be surprised.

Attempting to factor is a waste of time.

This:
"If it is more than 0 + delta(e) where delta(e) goes to zero"
doesn't even make sense. What do you mean by "goes to zero".

Aside from that, let's just say that the coefficients of a general quadratic equation have a uniform distribution across the reals (in text-book problems, I'd bet $1 Million that this is not the case) and that I just won't try to factor the quadratic into linear factors without using the QF. You are still missing the point that it is much easier for me (and I'd imagine for people who have been using the QF for more than 3 weeks) to look at the equation and determine without writting anything down what I need to put into my QF than it is to say "gee, ok, let me see, this is the first form which means I need to use this QF."

Again, your QF is simply a minor algebraic tweak (AS I PROVED) of the normal QF. You have done nothing.

 

[agentredlum]

That probability is 1. I can factor each quadratic equation.

Your turn: prove your previous assertion.

Uh-hmmmmm

I think you have some problems understanding the English language.

It is a waste of time answering your challenges, you will disagree anyway.

 

[micromass]

Uh-hmmmmm

I think you have some problems understanding the English language.

It is a waste of time answering your challenges, you will disagree anyway.

Wow, start to insult me? Way to prove your point!

 

[Robert1986]

May I ask a question? What math classes have you taken, agentredlum? Have you taken the Calc. sequence, yet?

 

[agentredlum]

This:
"If it is more than 0 + delta(e) where delta(e) goes to zero"
doesn't even make sense. What do you mean by "goes to zero".

Aside from that, let's just say that the coefficients of a general quadratic equation have a uniform distribution across the reals (in text-book problems, I'd bet $1 Million that this is not the case) and that I just won't try to factor the quadratic into linear factors without using the QF. You are still missing the point that it is much easier for me (and I'd imagine for people who have been using the QF for more than 3 weeks) to look at the equation and determine without writting anything down what I need to put into my QF than it is to say "gee, ok, let me see, this is the first form which means I need to use this QF."

Again, your QF is simply a minor algebraic tweak (AS I PROVED) of the normal QF. You have done nothing.

I posted the 'trick'

You keep claiming it's your proof

That's not honest.

 

[micromass]

I posted the 'trick'

You keep claiming it's your proof

That's not honest.

No, he's claiming that your trick is absolutely trivial and useless. He didn't claim that it's his proof.

 

[agentredlum]

May I ask a question? What math classes have you taken, agentredlum? Have you taken the Calc. sequence, yet?

I have completed calc 1, 2, 3 all A using Stewarts book

completed Linalg with A

Every math course i have taken, i did not get less than A

I worked 10 years as co-ordinator of math learning center at university.

Not only have i helped students but all tutors would come to me when they got stuck.

I have used the QF thousands of times, pen to paper so your comments about 3 weeks and 4th grade math are hurtfull

 

[Robert1986]

I posted the 'trick'

You keep claiming it's your proof

That's not honest.

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.

 

[agentredlum]

No, he's claiming that your trick is absolutely trivial and useless.

That's not the impression one gets when reading his posts.

Whether it's trivial or not is a matter of opinion.

I'm sure someone will find a use for it, especially in computer programming.

 

[micromass]

I have completed calc 1, 2, 3 all A using Stewarts book

completed Linalg with A

Every math course i have taken, i did not get less than A

I worked 10 years as co-ordinator of math learning center at university.

Not only have i helped students but all tutors would come to me when they got stuck.

I have used the QF thousands of times, pen to paper so your comments about 3 weeks and 4th grade math are hurtfull

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...

 

[Robert1986]

I have completed calc 1, 2, 3 all A using Stewarts book

completed Linalg with A

Every math course i have taken, i did not get less than A

I worked 10 years as co-ordinator of math learning center at university.

Not only have i helped students but all tutors would come to me when they got stuck.

I have used the QF thousands of times, pen to paper so your comments about 3 weeks and 4th grade math are hurtfull

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.

 

[agentredlum]

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.

 

[agentredlum]

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

 

[Robert1986]

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.

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.

 

[micromass]

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.

 

[Robert1986]

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.

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.

 

[agentredlum]

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.

 

[micromass]

You are using the word trivial to mean useless.

No, I didn't.

 

[agentredlum]

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

 

[agentredlum]

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?

 

[micromass]

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.

 

[Robert1986]

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.

 

[Robert1986]

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.

 

[agentredlum]

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.

 

[micromass]

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...

 

[Robert1986]

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.

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.

 

[agentredlum]

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!

 

[agentredlum]

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?

 

[micromass]

What does it mean to a computer, in terms of memory allocation?

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

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...

 

[Robert1986]

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.

 

[Borek]

This thread becomes trivial and useless. Locked.