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Math Help - Probability with relation to Quadratic Equation

  1. #1
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    {SOLVED} Probability with relation to Quadratic Equation

    First of all, I would like to thank each and everyone of you who is viewing this post. I know that you personally have helped many students with their questions and I know that by posting here, Im sure to get the right response !

    Heres the Question :


    3) Suppose we begin with the quadratic equation


    We know the roots of the equation are given by the quadratic formula,

    Now let a, b, and c be all positive single digit integers (that is, from 1 to 9). For example, one such quadratic equation can be (a=3, b=4, c=1)


    Of all the possible quadratic equations of this type, if one equation is randomly picked, what is the probability that the roots of the equation will be rational?



    Thank you in advance for all your help!
    hopefully someone can help me with this question !
    Last edited by somster100; July 12th 2011 at 01:23 AM.
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  2. #2
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    Re: Probability with relation to Quadratic Equation

    Quote Originally Posted by somster100 View Post
    First of all, I would like to thank each and everyone of you who is viewing this post. I know that you personally have helped many students with their questions and I know that by posting here, Im sure to get the right response !

    Heres the Question :


    3) Suppose we begin with the quadratic equation


    We know the roots of the equation are given by the quadratic formula,

    Now let a, b, and c be all positive single digit integers (that is, from 1 to 9). For example, one such quadratic equation can be (a=3, b=4, c=1)


    Of all the possible quadratic equations of this type, if one equation is randomly picked, what is the probability that the roots of the equation will be rational?


    Thank you in advance for all your help!
    hopefully someone can help me with this question !
    Click: Google
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  3. #3
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    Re: Probability with relation to Quadratic Equation

    Quote Originally Posted by mr fantastic View Post
    Click: Google
    Thank you for your help!
    So I narrowed down the question in being

    - What is the probability of b^2 - 4ac being a perfect square using intergers from 1-9

    Can anybody help me solve this please?
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  4. #4
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    Re: Probability with relation to Quadratic Equation

    Quote Originally Posted by somster100 View Post
    Thank you for your help!
    So I narrowed down the question in being

    - What is the probability of b^2 - 4ac being a perfect square using intergers from 1-9

    Can anybody help me solve this please?
    Can't you substitute b = 1 and see what must happen for a and c. Then substitute b = 2 and see what must happen for a and c etc. .... So you count the number of triplets (b, a, c) that satisfy the condition. And what is the total number of unrestricted triplets ....? Divide by those two numbers.
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    Re: Probability with relation to Quadratic Equation

    Quote Originally Posted by mr fantastic View Post
    Can't you substitute b = 1 and see what must happen for a and c. Then substitute b = 2 and see what must happen for a and c etc. .... So you count the number of triplets (b, a, c) that satisfy the condition. And what is the total number of unrestricted triplets ....? Divide by those two numbers.
    That works ! but why divide by those two numbers?
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  6. #6
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    Re: Probability with relation to Quadratic Equation

    Quote Originally Posted by somster100 View Post
    That works ! but why divide by those two numbers?
    Doesn't the question ask for a probability ...? And you will have learned that Pr(favourable outcome) = (Number of favourable outcomes)/(Total number outcomes) ....?
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  7. #7
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    This is taking wayyy to long to find each and every possibily -- # of faborable outcomes
    is there any other way of finding this?

    there must be another way for finding all the outcomes
    Last edited by mr fantastic; July 11th 2011 at 10:16 PM. Reason: Merged posts.
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    Re: Probability with relation to Quadratic Equation

    Quote Originally Posted by somster100 View Post
    there must be another way for finding all the outcomes
    The total number of triplets (b, a, c) where a, b and c are defined in post #1 is obviously 9^3.

    As for the favourable triplets, it surely doesn't take all that long to list them, especially since the discriminant can't be negative and the perfect square can't be larger than 81 ....
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