- #1
e.pramudita
- 14
- 0
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!