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anonymous
 4 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?
anonymous
 4 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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anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0ATTEMPTED 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?

mathslover
 4 years ago
Best ResponseYou've already chosen the best response.0you're right ... I think

phi
 4 years ago
Best ResponseYou've already chosen the best response.0I think you are allowed to use duplicates.

mathmate
 4 years ago
Best ResponseYou've already chosen the best response.0I think there is a typo in the answer, 48 instead of 88. For three _different_ numbers, the maximum cannot exceed 5*4*3=60.

mathmate
 4 years ago
Best ResponseYou've already chosen the best response.0Now, did you attempt the second part?

mathmate
 4 years ago
Best ResponseYou've already chosen the best response.0Ooooh, 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!

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0What? Please englighten me. Ywah I confused of the next part too.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Nahs. I don't think there is minus included. Because they'll list  number too if that wa the case

mathmate
 4 years ago
Best ResponseYou've already chosen the best response.0So we conclude that there is a typo in the answer. For part 2, you are familiar with the implications of the discriminant?

phi
 4 years ago
Best ResponseYou've already chosen the best response.0I don't see any way to get 88. If it turns out to be 88, I would like to know how.

mathmate
 4 years ago
Best ResponseYou've already chosen the best response.0Good, so out of the elements of the set, which element(s) put in "b" will give real zeroes?

mathmate
 4 years ago
Best ResponseYou've already chosen the best response.0or which are the combinations that will give real zeroes?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0There is so many combination I can't even begin to think

mathmate
 4 years ago
Best ResponseYou've already chosen the best response.0There is a maximum of 48, but many of them are obvious.

phi
 4 years ago
Best ResponseYou've already chosen the best response.0there are 48 combinations. brute force will work

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0There is at least 12 tho. Put c=0. so 4*3*1

mathmate
 4 years ago
Best ResponseYou've already chosen the best response.0Make 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.

mathmate
 4 years ago
Best ResponseYou've already chosen the best response.0I can start you off, assuming all the coefficients are positive, as agreed upon. Real 1,7,X X,7,1 Complex: X,0,X

mathmate
 4 years ago
Best ResponseYou've already chosen the best response.0Now add the number of cases after each line, and continue until you have a total of 48. Make sure that you don't overlap.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0i got 21 what did you get?

mathmate
 4 years ago
Best ResponseYou've already chosen the best response.0Do you mind showing your work?

mathmate
 4 years ago
Best ResponseYou've already chosen the best response.0I 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?

mathmate
 4 years ago
Best ResponseYou've already chosen the best response.0Sorry, 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} (93 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

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Really? I believe found more than 18 quadratic equations with imaginary roots.

mathmate
 4 years ago
Best ResponseYou've already chosen the best response.0Yes, I have 30 for complex, and 18 for real.

phi
 4 years ago
Best ResponseYou've already chosen the best response.0Here 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

mathmate
 4 years ago
Best ResponseYou've already chosen the best response.0Wow, 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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