Question

Find the value of k for which the given equation has real roots:
kx² - 6x - 2 = 0

Answer 1

The given equation is :
kx² - 6x - 2 = 0
Here, a = k, b = -6 & c = -2
This equation has real root if
b² - 4ac ≥ 0
⇒ (-6)² - 4 x k x (-2) ≥ 0
⇒ 36 + 8k ≥ 0
⇒ 8k ≥ -36
⇒ k ≥ - ${\displaystyle \frac{36}{8}}$
⇒ k ≥ ${\displaystyle -\frac{9}{2}}$.