Question

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?

Answer 1

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?

Answer 2

you're right ... I think

Answer 3

I think you are allowed to use duplicates.

Answer 4

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.

Answer 5

Now, did you attempt the second part?

Answer 6

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!

Answer 7

What? Please englighten me. Ywah I confused of the next part too.

Answer 8

@helder_edwin

Answer 9

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

Answer 10

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

Answer 11

D<0 imaginary

Answer 12

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

Answer 13

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

Answer 14

or which are the combinations that will give real zeroes?

Answer 15

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

Answer 16

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

Answer 17

there are 48 combinations. brute force will work

Answer 18

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

Answer 19

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.

Answer 20

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

Answer 21

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.

Answer 22

i got 21 what did you get?

Answer 23

Do you mind showing your work?

Answer 24

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?

Answer 25

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

Answer 26

Really? I believe found more than 18 quadratic equations with imaginary roots.

Answer 27

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

Answer 28

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

Answer 29

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! :)