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Question
Find the value of ‘k’ to identify the roots of the following equation is real and equal
kx2 + (6k + 2)x + 16 = 0
Solution
kx2 + (6k + 2)x + 16 = 0
a = k, b = (6k + 2), c = 16
Δ = b2 – 4ac ....[∵ the roots are real and equal]
⇒ (6k + 2)2 – 4 × k × 16 = 0
⇒ 36k2 + 24k + 4 – 64k = 0
⇒ 36k2 – 40k + 4 =0
⇒ 36k2 – 36k – 4k + 4 = 0
⇒ 36k (k – 1) – 4 (k – 1) = 0
⇒ 4 (k – 1) (9k – 1) = 0
⇒ k = 1 or k = `1/9`
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