What is the probability that the quadratic equation x^2 + 2bx + c = 0 has real roots?
I m not sure why the answer given was 8/10. Can some explanation be added so that i can understand the problem?
Thank you kindly.
- Dr DLv 71 decade agoFavorite Answer
We require 4b^2 - 4c > 0
or c < b^2
If c < 0, we are guaranteed to have real roots.
So let's choose b uniformly and randomly from the set of real numbers in (-N, N).
So the probaility of b being any value is db / 2N
Let's choose c from the set (-M, M).
c can take any value from -M to b^2, and we're safe.
So we expect M to be of the same order as N^2
P(reals | b) = (b^2 - -M) / 2M = (b^2 + M)/2M
P(reals) = ∑ P(reals | b) * P(b)
= ∫ (b^2 + M)/2M * db/(2N)
= 1/(4NM) * [b^3 /3 + Mb] ... b = -N to N
= 1/(2NM) * [N^3 /3 + NM]
= N^2 /(6M) + 1/2
This is where I'm stuck. If N^2 is of the same order as M, then you'll get
1/6 + 1/2 = 2/3
as M and N go to ∞
But don't quote me on this. This is generating quite a bit of discussion elsewhere.
- math guyLv 61 decade ago
The probability is 50%. We need the discriminant, b^2-4ac = b^2-4c, since a=1, to be non-negative in order to have real roots.
Now b^2 is never negative, so we need
Now, if we choose b, what fraction of all real numbers, c, would make this true? It doesn't matter what b is, because there are infinitely equally many numbers larger than (b^2)/4 as there are that are smaller than (b^2)/4, since the number line is infinite in size.
So there is an equal chance of having a positive discriminant as a negative one when b and c are randomly chosen. However, as soon as you start to limit how large (or small) b and c can be, then this probability changes.
I hope this helps!
For a more indepth analysis, read the responses at http://answers.yahoo.com/question/index;_ylt=Ar.C8...
- Anonymous1 decade ago
The probability that the quadratic equation x^2 + 2bx + c = 0 has two distinct roots is equal to the probability that its determinant (2b)^2-4*1*c is positive which in turn equals the probability that b^2>c. So the answer depends on the joint density of b and c.
If it happens that b and c are independent, their joint density is simply the product of their individual densities.
- DukeLv 71 decade ago
Assuming both random coefficients b and c to be independent and uniformly distributed, the answer is 1.
Please follow the link below and read Scarlet Manuka's and mine answers for explanation:
In Mosteller's book there is an important note: both questions about
x² + 2bx + c = 0 (Yours) and Ax² + Bx + C = 0 ARE NOT equivalent: if the coefficients A, B and C are independent and uniformly distributed in any cube, then B/A and C/A are neither independent, nor uniformly distributed.
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- 1 decade ago
This problem really does not make sense unless you are given some information about the probability distributions of b and c. You could assume something, but there is no basis given for making a reasonable assumption.
- 1 decade ago
- 1 decade ago
I don't think it can be answered simply or if it can be answered at all. You get infinities dividing infinities and I think it could involve some heavy number theory and cardinalities.