Question

1. Find the range of the values of k if kx^2 +8x > 6-k for all real values of x.
where k is positive​

Answer 1

Answer:

We have

kx2+8x+k<6⟺kx2+8x+k−6<0kx2+8x+k<6⟺kx2+8x+k−6<0

and this is always true when k<0k<0 and

b2−4ac=64−4k(k−6)<0⟹k2−6k−16>0b2−4ac=64−4k(k−6)<0⟹k2−6k−16>0

that is k<−2k<−2.

Hope it helps you.