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  • 3 years ago

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?

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  1. twitter
    • 3 years ago
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    ATTEMPTED SOLUTION: On question: "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?

  2. mathslover
    • 3 years ago
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    you're right ... I think

  3. phi
    • 3 years ago
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    I think you are allowed to use duplicates.

  4. mathmate
    • 3 years ago
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    I think there is a typo in the answer, 48 instead of 88. For three _different_ numbers, the maximum cannot exceed 5*4*3=60.

  5. mathmate
    • 3 years ago
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    Now, did you attempt the second part?

  6. mathmate
    • 3 years ago
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    Ooooh, as I started looking at the second part, some light goes on! What about the signs of the numbers? Each number (except for 0) has two signs!

  7. twitter
    • 3 years ago
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    What? Please englighten me. Ywah I confused of the next part too.

  8. twitter
    • 3 years ago
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    @helder_edwin

  9. twitter
    • 3 years ago
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    Nahs. I don't think there is minus included. Because they'll list - number too if that wa the case

  10. mathmate
    • 3 years ago
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    So we conclude that there is a typo in the answer. For part 2, you are familiar with the implications of the discriminant?

  11. twitter
    • 3 years ago
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    D<0 imaginary

  12. phi
    • 3 years ago
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    I don't see any way to get 88. If it turns out to be 88, I would like to know how.

  13. mathmate
    • 3 years ago
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    Good, so out of the elements of the set, which element(s) put in "b" will give real zeroes?

  14. mathmate
    • 3 years ago
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    or which are the combinations that will give real zeroes?

  15. twitter
    • 3 years ago
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    There is so many combination I can't even begin to think

  16. mathmate
    • 3 years ago
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    There is a maximum of 48, but many of them are obvious.

  17. phi
    • 3 years ago
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    there are 48 combinations. brute force will work

  18. twitter
    • 3 years ago
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    There is at least 12 tho. Put c=0. so 4*3*1

  19. mathmate
    • 3 years ago
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    Make two categories, enumerate them in one of the two categories. As phi said, brute force will work, and is probably the easiest way to do a simple case like this.

  20. mathmate
    • 3 years ago
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    I can start you off, assuming all the coefficients are positive, as agreed upon. Real 1,7,X X,7,1 Complex: X,0,X

  21. mathmate
    • 3 years ago
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    Now add the number of cases after each line, and continue until you have a total of 48. Make sure that you don't overlap.

  22. twitter
    • 3 years ago
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    i got 21 what did you get?

  23. mathmate
    • 3 years ago
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    Do you mind showing your work?

  24. mathmate
    • 3 years ago
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    I only get: X,X,0 (6 cases) 1,7,{3,5} (2 cases) {3,5},7,1 (2 cases) 1,5,3 3,5,1 for a total of 12 Did you find others?

  25. mathmate
    • 3 years ago
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    Sorry, it should have read: I only get: X,X,0 (12 cases) 1,7,{3,5} (2 cases) {3,5},7,1 (2 cases) 1,5,3 3,5,1 for a total of 18 For the complex roots, I have X,0,X (12 cases) [3,5,7},1,{3,5,7} (9 - 3 repeats = 6 cases) {1,5,7},3,{1,5,7} (9-3 repeats = 6 cases) 1,5,7 7,5,1 3,5,7 7,5,3 3,7,5 5,7,3 for a total of 30 cases

  26. twitter
    • 3 years ago
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    Really? I believe found more than 18 quadratic equations with imaginary roots.

  27. mathmate
    • 3 years ago
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    Yes, I have 30 for complex, and 18 for real.

  28. phi
    • 3 years ago
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    Here is the output of a short matlab program 1 1 0 3 0 2 1 0 5 0 3 1 0 7 0 4 1 3 0 1 5 1 3 5 0 6 1 3 7 0 7 1 5 0 1 8 1 5 3 1 9 1 5 7 0 10 1 7 0 1 11 1 7 3 1 12 1 7 5 1 13 3 0 1 0 14 3 0 5 0 15 3 0 7 0 16 3 1 0 1 17 3 1 5 0 18 3 1 7 0 19 3 5 0 1 20 3 5 1 1 21 3 5 7 0 22 3 7 0 1 23 3 7 1 1 24 3 7 5 0 25 5 0 1 0 26 5 0 3 0 27 5 0 7 0 28 5 1 0 1 29 5 1 3 0 30 5 1 7 0 31 5 3 0 1 32 5 3 1 0 33 5 3 7 0 34 5 7 0 1 35 5 7 1 1 36 5 7 3 0 37 7 0 1 0 38 7 0 3 0 39 7 0 5 0 40 7 1 0 1 41 7 1 3 0 42 7 1 5 0 43 7 3 0 1 44 7 3 1 0 45 7 3 5 0 46 7 5 0 1 47 7 5 1 0 48 7 5 3 0 18

  29. mathmate
    • 3 years ago
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    Wow, that's really brute force, ready to attack a bigger problem. I suppose the 18 at the bottom is the number of real solutions! :)

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