SOLUTION: Solve for the value of k that will make the roots of the function be real and equal: f(x)=(K-2)x^(2)+3Kx+3K+6

Algebra ->  Expressions -> SOLUTION: Solve for the value of k that will make the roots of the function be real and equal: f(x)=(K-2)x^(2)+3Kx+3K+6       Log On


   

Question 765365: Solve for the value of k that will make the roots of the function be real and equal: f(x)=(K-2)x^(2)+3Kx+3K+6
Answer by josgarithmetic(39617) About Me  (Show Source):
You can put this solution on YOUR website!
Exercise is for using the discriminant.

%28%283k%29%5E2-4%28k-2%29%283k%2B6%29%29, the discriminant.

%289k%5E2-4%283k%5E2-12%29%29
%289k%5E2-12k%5E2%2B48%29
%28-3k%5E2%2B48%29

You want the discriminant to be 0 to make f(x) have only one solution. This solution will be real.

-3k%5E2%2B48=0
3k%5E2=48
k%5E2=24%2A2%2F3=8%2A2=16
highlight%28k=-4%29 OR highlight%28k=4%29