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If roots of a quadratic equation 3y2 + ky + 12 = 0 are real and equal, then find the value of ‘k’ - Algebra

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प्रश्न

If roots of a quadratic equation 3y2 + ky + 12 = 0 are real and equal, then find the value of ‘k’

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उत्तर

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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अध्याय 2: Quadratic Equations - Q.2 (B)

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