(k 1)x2-6(k 1) 3(k 9)=0
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Since the roots are equal b2 - 4ac = 0
Here a = 1, b = -2(1 + 3k), c = 7(3 + 2k)
Hence [-2(1 + 3k)]2 = 4(1)[7(3 + 2k)]
4(1 + 3k)2 = 4[7(3 + 2k)]
1 + 6k + 9k2 = 21 + 14k
9k2 - 8k - 20 = 0
Factorize the above equation to get the value of k
Here a = 1, b = -2(1 + 3k), c = 7(3 + 2k)
Hence [-2(1 + 3k)]2 = 4(1)[7(3 + 2k)]
4(1 + 3k)2 = 4[7(3 + 2k)]
1 + 6k + 9k2 = 21 + 14k
9k2 - 8k - 20 = 0
Factorize the above equation to get the value of k
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