Question

In the following determine the set of values of k for which the given quadratic equation has real roots:

kx² + 6x + 1 = 0

Answer 1

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

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

= 6² - 4 x (k) x (1)

= 36 - 4k

The given equation will have real roots, if D ≥ 0

36 - 4k ≥ 0

4k ≤ 36

k ≤ 36/4

k ≤ 9

Therefore, the value of k ≤ 9.