Questions about quadratic formula

[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.