find the value of k in kx(x-2)+6=0
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Answer:
Given kx(x – 2) + 6 = 0
kx2 – 2kx + 6 = 0
Since the given quadratic equation has equal roots we have b2 – 4ac = 0
That is (– 2k)2 – 4(k)(6) = 0
⇒ 4k2 – 24k = 0
⇒ 4k(k – 6) = 0
⇒ 4k = 0 or (k – 6) = 0
∴ k = 0 or 6
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