## twitter 2 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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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

you're right ... I think

3. phi

I think you are allowed to use duplicates.

4. mathmate

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

Now, did you attempt the second part?

6. mathmate

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.

@helder_edwin

Nahs. I don't think there is minus included. Because they'll list - number too if that wa the case

10. mathmate

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

12. phi

I don't see any way to get 88. If it turns out to be 88, I would like to know how.

13. mathmate

Good, so out of the elements of the set, which element(s) put in "b" will give real zeroes?

14. mathmate

or which are the combinations that will give real zeroes?

There is so many combination I can't even begin to think

16. mathmate

There is a maximum of 48, but many of them are obvious.

17. phi

there are 48 combinations. brute force will work

There is at least 12 tho. Put c=0. so 4*3*1

19. mathmate

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

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

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?

23. mathmate

Do you mind showing your work?

24. mathmate

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

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.

27. mathmate

Yes, I have 30 for complex, and 18 for real.

28. 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

29. mathmate

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! :)