Helping my step-daughter with her homework?

Question

1. Find all values of k such that the equation: 3x(squared)-2x+k=0 has imaginary roots.

Answer 1

If you think back to the quadratic equation, you need the part that lives under the square root to be negative to have imaginary roots. The part that lives under the square root generally is b^2 - 4ac for ax^2 + bx +c. For your example it is 4 - 12k.
4-12k < 0
-12k < -4
k > 1/3

Answer 2

Your equation is:
3x² - 2x + k = 0

Using the coefficients (a = 3, b = -2, c = k), you can solve this with the quadratic equation which is:

...... -b ± sqrt( b² - 4ac )
x = ---------------------------
.................. 2a

You will get imaginary results when you are taking the square root of a negative number. So basically you just have to figure out when b² - 4ac (called the discriminant) is less than zero.

b² - 4ac < 0

Plug in your values for the coefficients: a = 3, b = -2, c = k

(-2)² - 4(3)(k) < 0

Simplify:
4 - 12k < 0

Subtract 4 from both sides:
-12k < -4

Multiply by -1. In this case you have to remember to reverse the direction of the sign because you are multiplying by a negative:
12k > 4

Divide both sides by 12:
k > 4/12
k > 1/3

So the answer is all values of k greater than 1/3 will result in two *imaginary* roots.

(Conversely, all values of k < 1/3 will result in two *real* roots. And if k = 1/3 you will have *one* real root).

Answer 3

3x^2 - 2x + k = 0

all quadratics can be written in the form ax^2 + bx + c

So, 3 is the a value, -2 is the b value, and k is the c value of this algebraic equation.

You can find if there are imaginary roots by solving the equation in the quadratic formula. You can do it even more simply by looking at the discriminant, which is part of the quadratic formula. If the discriminant, which is b^2 - 4ac, is negative, you have imaginary roots because you end up taking the square root of a negative value.

so, (-2)^2 - 4(3)(k) < 0 (meaning that its negative). simply solve this like a regular equation.

4 - 12k < 0
4 < 12k
k < (1/3)

So basically, as long as k is greater than 1/3, you will have imaginary roots.

Answer 4

If you put this into the quadratic equation you can find out

-B+/-(Sqrt)B^2-4AC all over 2A... You need to know when B^2-4AC is negative so B^2-4AC<0
2^2-4*3*K<0
4-12K<0
-12K<-4
K>-4/-12 (when dividing by a negative flip any inequality signs)
K>1/3

All values where K>1/3 would give imaginary roots

Answer 5

use formula b^2-4ac<0
(-2)^2-4(3)(k)<0
4-12k<0
k>1/3

so values of k are 1, 2, 3 and so on

Answer 6

Use the discriminant!
If you don't know what that is, you shouldn't have been given the homework