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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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?

Probability
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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?
you're right ... I think
  • phi
I think you are allowed to use duplicates.

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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.
Now, did you attempt the second part?
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!
What? Please englighten me. Ywah I confused of the next part too.
Nahs. I don't think there is minus included. Because they'll list - number too if that wa the case
So we conclude that there is a typo in the answer. For part 2, you are familiar with the implications of the discriminant?
D<0 imaginary
  • phi
I don't see any way to get 88. If it turns out to be 88, I would like to know how.
Good, so out of the elements of the set, which element(s) put in "b" will give real zeroes?
or which are the combinations that will give real zeroes?
There is so many combination I can't even begin to think
There is a maximum of 48, but many of them are obvious.
  • phi
there are 48 combinations. brute force will work
There is at least 12 tho. Put c=0. so 4*3*1
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.
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
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.
i got 21 what did you get?
Do you mind showing your work?
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?
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
Really? I believe found more than 18 quadratic equations with imaginary roots.
Yes, I have 30 for complex, and 18 for real.
  • phi
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
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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