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Permutation and Combination

  1. Jan 12, 2013 #1
    1. The problem statement, all variables and given/known data
    If three different number are taken from the set {0, 1, 3, 5, 7} to be used as the coefficient of a standard quadratic equation, how many such quadratic equations can be formed? How many of these have real roots?


    2. Relevant equations



    3. The attempt at a solution
    Part a: "How many such quadratic equations can be formed?"

    Quadratic equation is defined by ax^2 +bx +c =0 , a is not 0
    First box (x^2) = 4 (consist of everything but zero)
    Second box (x^1) = 4 (consist of everything but the number on the first box)
    Third box (x^0) = 3 (consist of everything but the number of the first box and the second box)
    So 4*4*3=48.
    But the answer says 88. Is the answer wrong?


    Part b: "How many of these have real roots?
    Imaginary roots have D<0
    So lets find how many equation that have imaginary roots and then subtract Part(a) answer with that number.

    Case 1: x^2 -> non zero -> (4)
    x^1 -> {0,1,3} -> (3)
    x^0 -> {non zero except the first and second box} -> (3)
    So 4*3*3=36

    Case 2: x^2 -> non zero -> (4)
    x^1 -> {5} -> (1)
    x^0 -> {3,7} ->(2)
    So 4*1*2=8

    I can think of 2 other cases but let stop for a second here.
    From here on we can conclude that there is at least 36+8=44 combination. Thus there are 48-44=4 quadratic equation that have real roots.

    But I can think of at least 12 real roots.
    Set coefficient x^0 = 0 and other coefficient with the rest.
    We can get.
    4*3*1=12

    But the answer is 28.

    I am confused!
     
  2. jcsd
  3. Jan 12, 2013 #2
    the standard quadratic equation is f(x) = a(x - h)2 + k
     
  4. Jan 12, 2013 #3
    Can you give more explanation on how to solve it?
     
  5. Jan 12, 2013 #4

    CAF123

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    Taking the quadratic to be of the form ##ax^2 + bx + c##, ##a \neq 0##, using a slightly different method to yours, I also get 48.
     
  6. Jan 12, 2013 #5
    IF the number that can be used is only 0, 1, 3, 5, 7, without any restriction, the numbers of equations that can be formed at most are only 5P3=60, right?
    So, do u think it means 10, 37, 137,... those numbers can be be used?
     
  7. Jan 12, 2013 #6

    CAF123

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    There is restriction in that a cannot be 0. I think it means three numbers from the given set( I.e you cannot combine them to form larger digit numbers)
     
  8. Jan 12, 2013 #7
    There is a restriction
     
  9. Jan 12, 2013 #8
    How about the second part?
     
  10. Jan 12, 2013 #9
    i mean if no restriction answer = 60, then when there is a restriction answer will be < 60...
     
  11. Jan 12, 2013 #10

    CAF123

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    For real root quadratics, you have ##b^2 - 4ac \geq 0\,\Rightarrow\,b^2 \geq 4ac## (valid because ##a,c ## both positive). How many combinations of a,b,c satisfy this?
     
    Last edited: Jan 12, 2013
  12. Jan 12, 2013 #11

    haruspex

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    For the second part, consider the cases in this order:
    c = 0
    c > 0, b = 0 or 1
    c > 0, b = 3
    c > 0, b > 3
     
  13. Jan 13, 2013 #12

    CAF123

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    For part a), note that the answer cannot possibly be 88. As thchian pointed out, there would be, without any restrictions, 5P3 = 60 possibilities so with some restrictions this number must be less than 60. To see that it is necessarily 48 again, count the number of combinations which result in a being 0. This gives 4P2 = 12. So subtracting gives 60 -12 = 48.
     
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