What are the possible values of k such that the equation 3x^2-2x+k=0 has imaginary roots?

Answer 1

this means that b^2-4AC must be less then zero
plug in the numbers
4-(4)(3)(k) less then 0
4-12k less then 0
1-3k less then zero
1 less then 3k
1/3 less then k
k greater then 1/3

Answer 2

In order for a quadradic to have imaginary roots, the determinate must be less than 0. This is because you take the square root of the determinate in the quadradic equation
[-b +/- SQRT(b^2-4ac)] / 2a

In your quadradic, a = 3, b = -2 and c = k

You are looking for solutions to the inequality:

4 - 4 * 3 * k < 0
4 - 12k < 0
4 < 12k
1/3 < k

So, K must be greater than 1/3 for the equation to have imaginary roots.

Answer 3

for any quadratic equation, ax^2 + bx + c = 0, the discriminant is b^2 - 4ac. whilst the discriminant is detrimental (<0) then you definitely could have imaginary roots 3x^2 - 2x + ok =0 a =3, b = -2 and c = ok b^2 - 4ac < 0 (-2)^2 - 4(3)(ok) < 0 4 - 12k < 0 -12k < -4 ok > a million/3 All values of ok are ok > a million/3 that provide you imaginary roots

Answer 4

k > 1/3 then (--2)^2 -- 4*3*k < 0 hence roots imaginary.

Answer 5

must have (-2)² - 4(3)k < 0
4 - 12k < 0
4 < 12k
1/3 < k