1. Sep 4, 2011

### 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?

2. Sep 4, 2011

### Dickfore

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

3. Sep 4, 2011

### agentredlum

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

4. Sep 4, 2011

### Dickfore

Define random in this context.

5. Sep 4, 2011

### agentredlum

Random real numbers for the coefficients a, b, c

6. Sep 4, 2011

### Dickfore

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

7. Sep 4, 2011

### disregardthat

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

8. Sep 4, 2011

How so?

9. Sep 4, 2011

### disregardthat

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

10. Sep 4, 2011

### agentredlum

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. Sep 4, 2011

### Dickfore

Then solve a problem in combinatorics.

12. Sep 4, 2011

### 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.

13. Sep 4, 2011

### disregardthat

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

14. Sep 4, 2011

### 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:

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15. Sep 4, 2011

### 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.

16. Sep 4, 2011

### agentredlum

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

17. Sep 4, 2011

### Dickfore

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

18. Sep 4, 2011

### jambaugh

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

19. Sep 4, 2011

### agentredlum

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. Sep 4, 2011

### agentredlum

Yes, this is what i am asking.