A) \[5\]
B) \[9\]
C) Both [a] and [b]
D) \[0\]
Correct Answer: C
Solution :
Given equation is \[{{x}^{2}}-(3k-1)x+2{{k}^{2}}+2k-11=0\] For equal roots \[{{b}^{2}}-4ac=0\] \[\Rightarrow \] \[{{(3k-1)}^{2}}=4(2{{k}^{2}}+2k-11)\] \[\Rightarrow \] \[9{{k}^{2}}+1-6k=8{{k}^{2}}+8k-44\] \[\Rightarrow \] \[{{k}^{2}}-14k+45=0\] \[\Rightarrow \] \[(k-5)(k-9)=0\] \[\Rightarrow \] \[k=5\] or \[k=9\]
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