3x^2+3x+k=0, find value(s) for no solution.

Question

find values for no solution and write answer as an inequality

3x^2+3x+k=0

Answer 1

There is always a solution; the question is whether the solution is real or complex.

 

By the quadratic formula,

 

x = [1 / 2(3)] * {-3 ±√[(3)^2 - 4(3)k]}

 

The real question is whether the discriminant allows real solutions or not.

 

D = (3)^2 - 4(3)k = 9 - 12k ≥ 0 to have real roots, < 0 to have complex roots.

 

9 ≥ 12k

3 ≥ 4k

 

So k ≤ 3/4 is a necessary condition to have real roots, otherwise the roots are complex.

Answer 2

By no solution, I assume you mean complex solutions.  You will have complex solutions if the discriminant is negative.

 

b² - 4ac < 0

 

 

where:

a = 3

b = 3

c = k

 

 

Plugging these value into the inequality I just set up for,

 

9 - 4(3)k < 0

 

9 - 12k < 0

 

   -12k < -9

 

        k > 9 / 12

 

        k > 3 / 4

 

Put this in decimal form for easier readability.

 

k > 0.75