What is the probability that the quadratic equation x^2 + 2bx + c = 0 has real roots?

Update:

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.

7 Answers

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  • Dr D
    Lv 7
    1 decade ago
    Favorite 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.

    http://answers.yahoo.com/question/index?qid=200803...

  • 1 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

    b^2-4c>0

    b^2>4c

    b^2/4>c

    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...

  • Anonymous
    1 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.

  • Duke
    Lv 7
    1 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:

    http://answers.yahoo.com/question/index?qid=200803...

    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.

  • 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.

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