A) 5
B) 9
C) Both (a) and (b)
D) 0
Correct Answer: C
Solution :
For the equation to be real roots, \[{{b}^{2}}-4ac=0\] Þ \[{{(3k-1)}^{2}}=4(2{{k}^{2}}+2k-11)\] Þ \[9{{k}^{2}}+1-6k=8{{k}^{2}}+8k-44\] Þ \[{{k}^{2}}-14k+45=0\] Þ \[(k-5)(k-9)\] = 0 \[k=5\]or\[k=9\].You need to login to perform this action.
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