Question

Find the values of k for which the given quadratic equation has real and distinct roots:

kx² + 6x + 1 = 0

Answer 1

The given quadric equation is kx² + 6x + 1 = 0, and roots are real and distinct.

Then find the value of k.

Here,

a = k, b = 6 and c = 1

As we know that D = b² - 4ac

Putting the value of a = k, b = 6 and c = 1

D = (6)² - 4 × (k) × (1)

= 36 - 4k

The given equation will have real and distinct roots, if D > 0

36 - 4k > 0

Now factorizing of the above equation

36 - 4k > 0

4k < 36

`k < 36/4`

k < 9

Now according to question, the value of k less than 9.

Therefore, the value of k < 9.