Question

Find the values of k for equation have real roots

kx(x – 3) + 9 = 0

Answer 1

Given,

kx(x – 3) + 9 = 0

It can be rewritten as,

kx² – 3kx + 9 = 0

It’s of the form of ax² + bx + c = 0

Where, a = k, b = -3k, c = 9

For the given quadratic equation to have real roots D = b² – 4ac ≥ 0

D = (-3k)² – 4(k)(9) ≥ 0

⇒ 9k² – 36k ≥ 0

⇒ 9k(k – 4) ≥ 0

⇒ k ≥ 0 and k ≥ 4

⇒ k ≥ 4

The value of k should be greater than or equal to 4 to have real roots.