The quadratic equation \((3k-2)x^2 + 12x + 3(k+1) = 0\) has equal roots. Find the two possible values of \(k\).
Answer (2)
x 1 = x 2 ⇒ Δ = 0 Δ = 1 2 2 − 4 ⋅ ( 3 k − 2 ) ⋅ 3 ( k + 1 ) Δ = 144 − ( 36 k − 24 ) ( k + 1 ) Δ = 144 − 36 k 2 − 36 k + 24 k + 24 Δ = − 36 k 2 − 12 k + 168 − 36 k 2 − 12 k + 168 = 0 − 3 k 2 − k + 14 = 0 − 3 k 2 + 6 k − 7 k + 14 = 0 − 3 k ( k − 2 ) − 7 ( k − 2 ) = 0 − ( 3 k + 7 ) ( k − 2 ) = 0 k = − 3 7 ∨ k = 2
The quadratic equation ( 3 k − 2 ) x 2 + 12 x + 3 ( k + 1 ) = 0 has equal roots for the values of k = 3 1 + 31 and k = 3 1 − 31 . To find these values, we set the discriminant equal to zero and solved the resulting quadratic equation in k. This gives us two possible solutions for k based on the conditions for equal roots.
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