Question

What is the smallest integral value of k such that 2x (kx -4) -x² + 6=0 has no real roots​

Answer 1

Answer:

K>19/3

Step-by-step explanation:

2x (kx -4) -x² + 6=0

2kx²-8x-x²+6=0

(2k-1)x²-8x+6=0

discriminant D =(8)²-4*6(2k-1)

=64-12k+24

=76-12k

For No real solutions

D<0

76-12k < 0

76< 12k

76/12 < k

19/3  < k

or k > 19/3