math grade 11
posted by priya on .
I need help with this question
determine the values in k for which the function f(x)= 4x^23x+2kx+1 has two zeros. Check these values in the original equation.
i don't get how to do it please help me.

F(x) = Y = 4x^2  3x + 2kx + 1.
This problem was solved by using EXCEL spread sheets and trial & error.
First, I temporarily ignored the 3rd term (2kx); and I changed b (the coefficient of x) until I found the required zeroes. Then I calculated the corresponding value of k:
b = 3 + 2k = 5,
3 + 2k = 5,
2k = 5 + 3,
2k = 8,
k = 8 / 2 = 4.
5 was chosen for b because it gave 2
zeroes.
So when k = 4: (1 , 0) , (1/4 , 0).