Question

find the value k for which quadratic equation kx2+6x+1=0 has real and distinct roots​

Answer 1

Answer:

in the quadratic equation Kx^2 - 6x - 1 = 0

a= K

b = -6

c = -1

since it does not have real roots

so

D < 0

b^2 - 4ac < 0

(-6)^2 - 4 × k(-1) < 0

36 + 4k < 0

36 + 4k - 36 < - 36

4k < -36

k < -9

so the value of k will be any number less than -9 .

Answer 2

Answer:

in real and distinct roots

D= b²-4ac

and D>0

6²-4k1

36-4k>0

36/4>k

k<9

hence k<9