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Question
If roots of a quadratic equation 3y2 + ky + 12 = 0 are real and equal, then find the value of ‘k’.
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Solution
3y2 + ky + 12 = 0
Comparing the above equation with
ax2 + by + c = 0, we get
a = 3, b = k, c = 12
∆ = b2 – 4ac
= (k)2 – 4 × 3 × 12
= k2 – 144
= k2 – (12)2
∆ = (k + 12)(k – 12) ...[∵ a2 – b2 = (a + b)(a – b)]
Since the roots are real and equal,
∆ = 0
∴ (k + 12)(k – 12) = 0
∴ k + 12 = 0 or k – 12 = 0
∴ k = –12 or k = 12
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