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Question
Find the value of ‘k’ to identify the roots of the following equation is real and equal
(5k – 6)x2 + 2kx + 1 = 0
Solution
(5k – 6)x2 + 2kx + 1 = 0
a = (5k – 6), b = 2k, c = 1
Δ = b2 – 4ac
⇒ (2k)2 – 4(5k – 6)(1)
⇒ 4k2 – 20k + 24 = 0 ...[∵ Since the roots are real and equal]
⇒ k2 – 5k + 6 = 0
⇒ (k – 3)(k – 2) = 0
k = 3, 2
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