How Many Quadratic Equations Can Be Formed With Real Roots?

[e.pramudita]

Homework Statement

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?

Homework Equations

The Attempt at a Solution

Part a: "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?Part b: "How many of these have real roots?
Imaginary roots have D<0
So let's find how many equation that have imaginary roots and then subtract Part(a) answer with that number.

Case 1: x^2 -> non zero -> (4)
x^1 -> {0,1,3} -> (3)
x^0 -> {non zero except the first and second box} -> (3)
So 4*3*3=36

Case 2: x^2 -> non zero -> (4)
x^1 -> {5} -> (1)
x^0 -> {3,7} ->(2)
So 4*1*2=8

I can think of 2 other cases but let stop for a second here.
From here on we can conclude that there is at least 36+8=44 combination. Thus there are 48-44=4 quadratic equation that have real roots.

But I can think of at least 12 real roots.
Set coefficient x^0 = 0 and other coefficient with the rest.
We can get.
4*3*1=12

But the answer is 28.

I am confused!

 

[thchian]

the standard quadratic equation is f(x) = a(x - h)2 + k

 

[e.pramudita]

the standard quadratic equation is f(x) = a(x - h)2 + k

Can you give more explanation on how to solve it?

 

[CAF123]

Taking the quadratic to be of the form $ax^2 + bx + c$, $a \neq 0$, using a slightly different method to yours, I also get 48.

 

[thchian]

Can you give more explanation on how to solve it?

IF the number that can be used is only 0, 1, 3, 5, 7, without any restriction, the numbers of equations that can be formed at most are only 5P3=60, right?
So, do u think it means 10, 37, 137,... those numbers can be be used?

 

[CAF123]

IF the number that can be used is only 0, 1, 3, 5, 7, without any restriction, the numbers of equations that can be formed at most are only 5P3=60, right?
So, do u think it means 10, 37, 137,... those numbers can be be used?

There is restriction in that a cannot be 0. I think it means three numbers from the given set( I.e you cannot combine them to form larger digit numbers)

 

[e.pramudita]

IF the number that can be used is only 0, 1, 3, 5, 7, without any restriction, the numbers of equations that can be formed at most are only 5P3=60, right?
So, do u think it means 10, 37, 137,... those numbers can be be used?

There is a restriction

If three different number are taken from the set

 

[e.pramudita]

Taking the quadratic to be of the form $ax^2 + bx + c$, $a \neq 0$, using a slightly different method to yours, I also get 48.

How about the second part?

 

[thchian]

There is a restriction

i mean if no restriction answer = 60, then when there is a restriction answer will be < 60...

 

[CAF123]

How about the second part?

For real root quadratics, you have $b^2 - 4ac \geq 0\,\Rightarrow\,b^2 \geq 4ac$ (valid because $a,c$ both positive). How many combinations of a,b,c satisfy this?

 

[haruspex]

For the second part, consider the cases in this order:
c = 0
c > 0, b = 0 or 1
c > 0, b = 3
c > 0, b > 3

 

[CAF123]

For part a), note that the answer cannot possibly be 88. As thchian pointed out, there would be, without any restrictions, 5P3 = 60 possibilities so with some restrictions this number must be less than 60. To see that it is necessarily 48 again, count the number of combinations which result in a being 0. This gives 4P2 = 12. So subtracting gives 60 -12 = 48.