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Question
Find the value of k for which the given equation has real roots:
kx2 - 6x - 2 = 0
Solution
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 ≥ - `(36)/(8)`
⇒ k ≥ `-(9)/(2)`.
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