SOLUTION: f(x) = x2 - kx + 9, where k is a constant find the set of values of k for which the equation f(x) = 0 has no real solutions I know that b2 - 4ac<0 has no real solutions

Algebra ->  Functions -> SOLUTION: f(x) = x2 - kx + 9, where k is a constant find the set of values of k for which the equation f(x) = 0 has no real solutions I know that b2 - 4ac<0 has no real solutions      Log On


   

Question 56052: f(x) = x2 - kx + 9, where k is a constant
find the set of values of k for which the equation f(x) = 0 has no real solutions
I know that b2 - 4ac<0 has no real solutions
Answer by venugopalramana(3286) About Me  (Show Source):
You can put this solution on YOUR website!
f(x) = x2 - kx + 9, where k is a constant
find the set of values of k for which the equation f(x) = 0 has no real solutions
I know that b2 - 4ac<0 has no real solutions
CORRECT SO FIND B^2-4AC...AND......MAKE IT < 0
B^2-4AC = K^2-4*1*9=K^2-36<0
K^2<36
|K|< 6..THAT IS K LIES BETWEEN -6 AND +6.....(-6,6)...