Questions about quadratic formula

In summary, the conversation discusses the concept of random second-degree equations and how to determine the probability of computing double or triple negatives for the coefficients -b and -4ac in the formula x = (-b +-sqrt(b^2 - 4ac))/(2a). The conversation also delves into the difficulties of defining a random polynomial of second degree due to the infinite set of real numbers and the need to specify a probability distribution for the coefficients. A possible solution is to consider regions of types by plotting the parameters B and C in the Cartesian plane.
  • #1
agentredlum
484
0
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?
 
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  • #2
LOL, what are you talking about? 'Percent of equations'?!
 
  • #3
Dickfore said:
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?
 
  • #4
Define random in this context.
 
  • #5
Dickfore said:
Define random in this context.

Random real numbers for the coefficients a, b, c
 
  • #6
'Random real number' is a contradictio in adjecto. If you meant they are random variables, you must say something about their distribution.
 
  • #7
You will probably get into difficulties defining a random polynomial of second degree the way you want.
 
  • #8
disregardthat said:
You will probably get into difficulties defining a random polynomial of second degree the way you want.

How so?
 
  • #9
agentredlum said:
How so?

You tell me, what is your distribution of the random variables a, b and c?
 
  • #10
disregardthat said:
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.
 
  • #11
Then solve a problem in combinatorics.
 
  • #12
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.
 
  • #13
There are infinitely many real numbers, it is nothing like a coinflip.
 
  • #14
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 [itex] ax^2 + bx + c = 0[/itex] 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 [itex] x^2 + Bx + C = 0[/itex] where B=b/a and C = c/a.
The solutions are: [itex] x = \frac{-B \pm \sqrt{B^2 - 4C}}{2}[/itex]
Now consider the range of solutions by plotting the points (B,C) in the Cartesian plane.

First graph the curve [itex] B^2 - 4C = 0[/itex] or [itex] C = \frac{1}{4}B^2[/itex] 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 [itex] -B = \pm\sqrt{B^2 - 4C}[/itex] or squaring the equation where:
[itex] B^2 = B^2 - 4C[/itex], or where C = 0 (the B axis).

Testing points between these curves you get the full range of cases. Here is a picture:
attachment.php?attachmentid=38547&stc=1&d=1315139361.gif
 

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  • #15
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.
 
  • #16
HallsofIvy said:
Until you specify a probability distribution for your coefficients, your question has no answer.

How do i do that? Can you provide some examples?
 
  • #17
jambaugh said:
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 [itex] ax^2 + bx + c = 0[/itex] 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 [itex] x^2 + Bx + C = 0[/itex] where B=b/a and C = c/a.
The solutions are: [itex] x = \frac{-B \pm \sqrt{B^2 - 4C}}{2}[/itex]
Now consider the range of solutions by plotting the points (B,C) in the Cartesian plane.

First graph the curve [itex] B^2 - 4C = 0[/itex] or [itex] C = \frac{1}{4}B^2[/itex] 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 [itex] -B = \pm\sqrt{B^2 - 4C}[/itex] or squaring the equation where:
[itex] B^2 = B^2 - 4C[/itex], or where C = 0 (the B axis).

Testing points between these curves you get the full range of cases. Here is a picture:
attachment.php?attachmentid=38547&stc=1&d=1315139361.gif

This is not what the OP had asked. He is only interested in 'double' and 'triple' negative occurrences as defined in his op.
 
  • #18
Dickfore said:
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!
 
  • #19
jambaugh said:
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 [itex] ax^2 + bx + c = 0[/itex] 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 [itex] x^2 + Bx + C = 0[/itex] where B=b/a and C = c/a.
The solutions are: [itex] x = \frac{-B \pm \sqrt{B^2 - 4C}}{2}[/itex]
Now consider the range of solutions by plotting the points (B,C) in the Cartesian plane.

First graph the curve [itex] B^2 - 4C = 0[/itex] or [itex] C = \frac{1}{4}B^2[/itex] 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 [itex] -B = \pm\sqrt{B^2 - 4C}[/itex] or squaring the equation where:
[itex] B^2 = B^2 - 4C[/itex], or where C = 0 (the B axis).

Testing points between these curves you get the full range of cases. Here is a picture:
attachment.php?attachmentid=38547&stc=1&d=1315139361.gif

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.
 
  • #20
Dickfore said:
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.
 
  • #21
agentredlum said:
Yes, this is what i am asking.

Cool, now provide your attempt at a solution.
 
  • #22
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.

agentredlum said:
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%.
 
  • #23
These are correct solutions.
 
  • #24
agentredlum said:
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]
 
  • #25
Dickfore said:
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.
 
Last edited:
  • #26
agentredlum said:
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 :smile:
 
  • #27
It would really help if you could extend your work to the http://planetmath.org/encyclopedia/QuarticFormula.html [Broken].
 
Last edited by a moderator:
  • #28
And I assume that the point of this is to show us that your "trick" is worthwhile?
 
  • #29
So 5 out of 8 times you are performing an un-necessary operation, because you have been taught a myth.

That's 62.5%
 
  • #30
agentredlum said:
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.
 
  • #31
agentredlum said:
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.
 
  • #32
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.
 
  • #33
agentredlum said:
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.
 
  • #34
micromass said:
Factoring is equivalent to solving the quadratic formula.

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

Factoring is not practical.
 
  • #35
agentredlum said:
Be that as it may, that's neither here nor there.

Factoring is not practical.

Proof for that, please?
 
<h2>What is the quadratic formula?</h2><p>The quadratic formula is a mathematical formula used to solve quadratic equations of the form ax^2 + bx + c = 0, where a, b, and c are constants.</p><h2>When do you use the quadratic formula?</h2><p>The quadratic formula is used when solving quadratic equations that cannot be easily factored or solved using other methods.</p><h2>How do you use the quadratic formula?</h2><p>To use the quadratic formula, you first need to identify the values of a, b, and c in the equation ax^2 + bx + c = 0. Then, plug these values into the formula: x = (-b ± √(b^2 - 4ac)) / 2a. Finally, solve for the two possible values of x.</p><h2>What does the discriminant represent in the quadratic formula?</h2><p>The discriminant, b^2 - 4ac, is the part of the quadratic formula under the square root sign. It represents the nature of the solutions to the quadratic equation. If the discriminant is positive, there are two real solutions. If it is zero, there is one real solution. And if it is negative, there are no real solutions, but two complex solutions.</p><h2>Are there any other methods for solving quadratic equations?</h2><p>Yes, there are other methods for solving quadratic equations such as factoring, completing the square, and graphing. However, the quadratic formula is a reliable method that can be used for any quadratic equation, regardless of its complexity.</p>

What is the quadratic formula?

The quadratic formula is a mathematical formula used to solve quadratic equations of the form ax^2 + bx + c = 0, where a, b, and c are constants.

When do you use the quadratic formula?

The quadratic formula is used when solving quadratic equations that cannot be easily factored or solved using other methods.

How do you use the quadratic formula?

To use the quadratic formula, you first need to identify the values of a, b, and c in the equation ax^2 + bx + c = 0. Then, plug these values into the formula: x = (-b ± √(b^2 - 4ac)) / 2a. Finally, solve for the two possible values of x.

What does the discriminant represent in the quadratic formula?

The discriminant, b^2 - 4ac, is the part of the quadratic formula under the square root sign. It represents the nature of the solutions to the quadratic equation. If the discriminant is positive, there are two real solutions. If it is zero, there is one real solution. And if it is negative, there are no real solutions, but two complex solutions.

Are there any other methods for solving quadratic equations?

Yes, there are other methods for solving quadratic equations such as factoring, completing the square, and graphing. However, the quadratic formula is a reliable method that can be used for any quadratic equation, regardless of its complexity.

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