Questions about quadratic formula

[agentredlum]

I have 2 questions about -b and - 4ac in the formula...

x = (-b +-sqrt(b^2 - 4ac))/(2a)

If you are given 2nd degree equations at random (that can be solved using above formula)

Question1: What percent of the equations given would you expect to compute a double negative for -b?

Question2: What percent of the equations given would you expect to compute a triple negative for - 4ac?

For instance if b is -5 then -(-5) must be evaluated, how often would you expect to see 2 negatives?

If a is -2 and c is -3 then - 4(-2)(-3) must be evaluated, how often would you expect to see 3 negatives?

 

[Dickfore]

LOL, what are you talking about? 'Percent of equations'?!

 

[agentredlum]

LOL, what are you talking about? 'Percent of equations'?!

Suppose you are given 1000 random equations, how many of them would require computing double negatives? How many would require computing triple negatives?

 

[Dickfore]

Define random in this context.

 

[agentredlum]

Define random in this context.

Random real numbers for the coefficients a, b, c

 

[Dickfore]

'Random real number' is a contradictio in adjecto. If you meant they are random variables, you must say something about their distribution.

 

[disregardthat]

You will probably get into difficulties defining a random polynomial of second degree the way you want.

 

[agentredlum]

You will probably get into difficulties defining a random polynomial of second degree the way you want.

How so?

 

[disregardthat]

How so?

You tell me, what is your distribution of the random variables a, b and c?

 

[agentredlum]

You tell me, what is your distribution of the random variables a, b and c?

I just want to explore the signs of a, b, c. The signs should have equal probability distribution, like flipping a fair coin for each sign, say tails=negative, heads=positive.

 

[Dickfore]

Then solve a problem in combinatorics.

 

[HallsofIvy]

Well, there's the problem. If you have only a finite number of possible outcomes, then it is perfectly reasonable to say "random" meaning "equally likely" (well, maybe not "reasonable" but commonly done!), the "uniform" distribution. However, the set of real numbers in any interval is not finite and there is no "equally likely" or "uniform" distribution. Until you specify a probability distribution for your coefficients, your question has no answer.

 

[disregardthat]

There are infinitely many real numbers, it is nothing like a coinflip.

 

[jambaugh]

Dickfore,
Don't be ungracious.

agentredlum,
The range of possible equations is an infinite set so you cannot really talk about percentages. You can however consider regions of types by plotting the parameters.

First consider the original equation ax^2 + bx + c = 0 and notice that multiplying the equation by a constant doesn't change its solutions. So really you have two free parameters defining which equation (and likewise two solutions).
Consider the equation of the form x^2 + Bx + C = 0 where B=b/a and C = c/a.
The solutions are: x = \frac{-B \pm \sqrt{B^2 - 4C}}{2}
Now consider the range of solutions by plotting the points (B,C) in the Cartesian plane.

First graph the curve B^2 - 4C = 0 or C = \frac{1}{4}B^2 which is a parabola with vertex (0,0) and passing through the points (2,1) and (-2,1).

• Above this parabola C is too big so you have no real solutions.
• Along the parabola the discriminant is 0 so you have one repeated solution.
• Below the parabola you have two distinct real solutions.

As to the signs of the solutions start by considering the case of one solution. Notice that with zero radical the sign of the solution is opposite the sign of B so cut the parabola in two halves the left half for positive repeated solutions and the right half for negative repeated solutions.

For the region below the parabola where there are two distinct solutions you have to consider the boundary case where one of the two solutions is zero.
That occurs when -B = \pm\sqrt{B^2 - 4C} or squaring the equation where:
B^2 = B^2 - 4C, or where C = 0 (the B axis).

Testing points between these curves you get the full range of cases. Here is a picture:

 

[Dickfore]

He is saying that a, b and c can have a probability of 1/2 for either sign independently of one another if I am not mistaken. The problem becomes that of elementary probability theory involving combinatorics.

 

[agentredlum]

Until you specify a probability distribution for your coefficients, your question has no answer.

How do i do that? Can you provide some examples?

 

[Dickfore]

Dickfore,
Don't be ungracious.

agentredlum,
The range of possible equations is an infinite set so you cannot really talk about percentages. You can however consider regions of types by plotting the parameters.

First consider the original equation ax^2 + bx + c = 0 and notice that multiplying the equation by a constant doesn't change its solutions. So really you have two free parameters defining which equation (and likewise two solutions).
Consider the equation of the form x^2 + Bx + C = 0 where B=b/a and C = c/a.
The solutions are: x = \frac{-B \pm \sqrt{B^2 - 4C}}{2}
Now consider the range of solutions by plotting the points (B,C) in the Cartesian plane.

First graph the curve B^2 - 4C = 0 or C = \frac{1}{4}B^2 which is a parabola with vertex (0,0) and passing through the points (2,1) and (-2,1).

• Above this parabola C is too big so you have no real solutions.
• Along the parabola the discriminant is 0 so you have one repeated solution.
• Below the parabola you have two distinct real solutions.

As to the signs of the solutions start by considering the case of one solution. Notice that with zero radical the sign of the solution is opposite the sign of B so cut the parabola in two halves the left half for positive repeated solutions and the right half for negative repeated solutions.

For the region below the parabola where there are two distinct solutions you have to consider the boundary case where one of the two solutions is zero.
That occurs when -B = \pm\sqrt{B^2 - 4C} or squaring the equation where:
B^2 = B^2 - 4C, or where C = 0 (the B axis).

Testing points between these curves you get the full range of cases. Here is a picture:

This is not what the OP had asked. He is only interested in 'double' and 'triple' negative occurrences as defined in his op.

 

[jambaugh]

This is not what the OP had asked. He is only interested in 'double' and 'triple' negative occurrences as defined in his op.

Oops! All that work with the graph and I should have read the post more carefully. My Bad!

 

[agentredlum]

Dickfore,
Don't be ungracious.

agentredlum,
The range of possible equations is an infinite set so you cannot really talk about percentages. You can however consider regions of types by plotting the parameters.

First consider the original equation ax^2 + bx + c = 0 and notice that multiplying the equation by a constant doesn't change its solutions. So really you have two free parameters defining which equation (and likewise two solutions).
Consider the equation of the form x^2 + Bx + C = 0 where B=b/a and C = c/a.
The solutions are: x = \frac{-B \pm \sqrt{B^2 - 4C}}{2}
Now consider the range of solutions by plotting the points (B,C) in the Cartesian plane.

First graph the curve B^2 - 4C = 0 or C = \frac{1}{4}B^2 which is a parabola with vertex (0,0) and passing through the points (2,1) and (-2,1).

• Above this parabola C is too big so you have no real solutions.
• Along the parabola the discriminant is 0 so you have one repeated solution.
• Below the parabola you have two distinct real solutions.

As to the signs of the solutions start by considering the case of one solution. Notice that with zero radical the sign of the solution is opposite the sign of B so cut the parabola in two halves the left half for positive repeated solutions and the right half for negative repeated solutions.

For the region below the parabola where there are two distinct solutions you have to consider the boundary case where one of the two solutions is zero.
That occurs when -B = \pm\sqrt{B^2 - 4C} or squaring the equation where:
B^2 = B^2 - 4C, or where C = 0 (the B axis).

Testing points between these curves you get the full range of cases. Here is a picture:

This is an amazing post. How did you construct it so quickly? Theres a lot of useful information in your post
jambaugh, I'm not sure how it is going to answer my questions?

I am interested in the signs of a, b, c, not in the nature of the solutions.

EDIT: Didn't see dickfore post.

 

[agentredlum]

He is saying that a, b and c can have a probability of 1/2 for either sign independently of one another if I am not mistaken. The problem becomes that of elementary probability theory involving combinatorics.

Yes, this is what i am asking.

 

[Dickfore]

Yes, this is what i am asking.

Cool, now provide your attempt at a solution.

 

[pwsnafu]

Well if we are only interested in the sign of the parameters, there are eight combinations in total: a can positive or negative, b can be positive or negative, and c can be positive or negative.

Question1: What percent of the equations given would you expect to compute a double negative for -b?

Well this is easy. b has a 50% per cent chance of being negative. We don't need to consider the other parameters.

Question2: What percent of the equations given would you expect to compute a triple negative for - 4ac?

So we need a negative and c negative. b can be either. So 2 / 8 = 25%.

 

[Dickfore]

These are correct solutions.

 

[sankalpmittal]

Yes, this is what i am asking.

Hiii agentredlum !

Yes b will have 0.5 probability to be either negative or positive .
Triple negatives occurrence i.e. (-4)(-a)(-c) will have 0.25 probability .

So , I think that all you are doing is to prove your formula in thread : https://www.physicsforums.com/showthread.php?t=510105 to be simplified version of treating such situations , right ?


[BTW see your last private message which I had sent]

 

[agentredlum]

Cool, now provide your attempt at a solution.

O-K, it's not a homework question but i will show my attempt as requested.

There are 3 positions where signs can change so there are a total of 8 possible configurations listed below
a, b, c
a, b, -c
a, -b, c
a, -b, -c
-a, b, c
-a, b, -c
-a, -b, c
-a, -b, -c

4 of these have -b so i would expect to compute a double negative, -(-b), half of the time.

2 of them have -a, -c so i would expect to compute a triple negative, -4(-a)(-c), a quarter of the time.

EDIT:Didn't see other posts.

 

[micromass]

4 of these have -b so i would expect to compute a double negative, -(-b), half of the time.

2 of them have -a, -c so i would expect to compute a triple negative, -4ac, a quarter of the time.

Seems correct

 

[jhae2.718]

It would really help if you could extend your work to the http://planetmath.org/encyclopedia/QuarticFormula.html .

 

[Robert1986]

And I assume that the point of this is to show us that your "trick" is worthwhile?

 

[agentredlum]

So 5 out of 8 times you are performing an un-necessary operation, because you have been taught a myth.

That's 62.5%

 

[micromass]

So 5 out of 8 times you are performing an un-necessary operation, because you have been taught a myth.

That's 62.5%

Not necessarily. You have assumed that the distribution of the coefficients is uniform. This doiesn't need to be the case in practise!
You need more data to decide that.

 

[Robert1986]

So 5 out of 8 times you are performing an un-necessary operation, because you have been taught a myth.

That's 62.5%

3 Things:

1) No one was taught a myth. The point of the QF is to have one formula that works for all quadratic equations, not just for one "form". No myth has been taught. Your forumula is (as I proved) a simple algebraic manipulation.

2) You are ignoring the fact that before I even think about using any QF, I'm going to try to factor it into two linear factors.

3) Those of us who have been using the QF for more than, say, 3 weeks don't actually write out explicitly the steps that you assume we do (at least, I am speaking for myself.) I see the quadratic and I know immeaditley what I need to plug into the QF to solve for x.

 

[agentredlum]

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.

 

[micromass]

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 + e where delta(e) goes to zero, I would be surprised.

Attempting to factor is a waste of time.

Factoring is equivalent to solving the quadratic formula. Your point is moot.

 

[agentredlum]

Factoring is equivalent to solving the quadratic formula.

Be that as it may, that's neither here nor there.

Factoring is not practical.

 

[micromass]

Be that as it may, that's neither here nor there.

Factoring is not practical.

Proof for that, please?

 

[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.