- #1

discombobulated

- 41

- 0

1) Find the set of values of k for which the equation: f(x)=x

^{2}-kx+16 has no real solutions.

This is what I've done:

discriminant < 0

k

^{2}-4(16) =0

K

^{2}-64=0

k=8

x

^{2}-8x+16

(x-4)(x-4)=0

x=4

But, i don't know what to do next now! Please help!

2) The quadratic equation kx

^{2}+(4k+1)x +(3k+1)=0 has a repeated root. find the value of k.

I'm not at all sure about this one, but i tried to do something anyway:

(4kx+x)

^{2}-12k

^{2}+4k=0

4kx +x -10k = 0

(2k+x)(2x-5) =0

I have no idea what I'm doing, please help!

Thanks!