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Question
If the roots of the given quadratic equation are real and equal, then find the value of ‘k’
kx(x – 2) + 6 = 0
Solution
kx(x – 2) + 6 = 0
∴ kx2 – 2kx + 6 = 0
Comparing the above equation with
ax2 + bx + c = 0, we get
a = k, b = – 2k, c = 6
∆ = b2 – 4ac
= (–2k)2 – 4 × k × 6
= 4k2 – 24k
∴ ∆ = 4k(k – 6)
Since the roots are real and equal,
∆ = 0
∴ 4k(k – 6) = 0
∴ k(k – 6) = 0
∴ k = 0 or k – 6 = 0
But, if k = 0, then quadratic coefficient becomes zero.
∴ k ≠ 0
∴ k = 6
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