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Math Help - Square equation problem

  1. #1
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    Square equation problem

    Hello
    I have equation:
    Code:
    x^2 - px + q = 0
    Equation has 2 solutions. p and q are integer and positive. How many different p and q exsist if both solutions are less then 5 (x1,x2<5).
    Sorry for my English. Please help. Cheers.
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  2. #2
    Super Member
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    Quote Originally Posted by matt12345 View Post
    Hello
    I have equation:
    Code:
    x^2 - px + q = 0
    Equation has 2 solutions. p and q are integer and positive. How many different p and q exsist if both solutions are less then 5 (x1,x2<5).
    Sorry for my English. Please help. Cheers.
    I do not understand this (x1,x2<5).
    Could you explain what that means in this situation?

    Since p & q are integers, x must be greater than zero, and an integer.
    Code:
     
    x  p  q
    1  1  0
    1  2  1
    1  3  2
    1  4  3
    2  2  0
    2  3  2
    2  4  4
    3  4  3
    That's eight solutions.
    If you removed x=1, p=1, q=0 and x=2, p=2, q=0, you would still have 6 solutions.

    Could you explain the restrictions so that only 2 work?
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  3. #3
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    Quote Originally Posted by matt12345 View Post
    Hello
    I have equation:
    Code:
    x^2 - px + q = 0
    Equation has 2 roots. p and q are integer and positive. How many different p and q exsist if both roots are less then 5 (x1,x2<5).
    Sorry for my English. Please help. Cheers.
    x1 and x2 are roots
    Root (mathematics) - Wikipedia, the free encyclopedia

    Sorry for my English
    Last edited by matt12345; October 1st 2009 at 08:49 AM.
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  4. #4
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    ref
    please help
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  5. #5
    Senior Member pankaj's Avatar
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    New Delhi(India)
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    Condition 1:

    Since roots are real and distinct:

    Discriminant>0

    p^2-4q>0

    4q<p^2................................(1)


    Condition 2:

    Since both roots are less than 5,the sum of the roots will be less than 10.

    p<10

    Condition 3:
    Let f(x)=x^2-px+q

    f(5)>0

    25-5p+q>0

    From conditions 1,2 and 3 we can see that 0<p<10 and 0<q<25,where p and q are integers

    I hope this helps.
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  6. #6
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    THX

    Why and what does this give? Why q<25 ?
    What will be the resualt?
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  7. #7
    Senior Member pankaj's Avatar
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    Since x_{1} and x_{2} are roots of f(x)=0,so (x-x_{1}) and (x-x_{2}) are factors of f(x)=0

    f(x)=x^2-px+q=(x-x_{1})(x-x_{2})

    Since both the roots are less than 5,clearly both (5-x_{1}) and (5-x_{2}) are positive.Thus

    f(5)=5^2-p(5)+q=(5-x_{1})(5-x_{2})>0

    Since 4q<p^2

    0<p<10

    \therefore 4q<p^2<100

    due to which 0<q<25

    Now these values of p and q must also satisfy the condition f(5)>0

    5p<25+q<25+25

    p<10.

    Thus the values 0<p<10 and 0<q<25 also satisfy the condition f(5)>0
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  8. #8
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    Thx
    But I don't know how many p&q satisfy the condition x^2 - px + q = 0 ?
    For exaple in this reason result is 16 (I have written program in c++ and check a lot of p & q, if condition was satisfied r++ and r was a resault).
    How can I get that?
    Last edited by matt12345; October 22nd 2009 at 01:25 PM.
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  9. #9
    Senior Member pankaj's Avatar
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    I do not understand your querry
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