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J & K CET Engineering J and K - CET Engineering Solved Paper-2008

  • question_answer
    A value of k for which the quadratic equation   \[{{x}^{2}}-2x(1+3k)+7(2k+3)=0\]has   equal roots, is

    A)  \[1\]               

    B)  \[2\]

    C)  \[3\]

    D)  \[4\]

    Correct Answer: B

    Solution :

    Given equation is \[{{x}^{2}}-2x(1+3k)+7(2k+3)=0.\] For equal roots, discriminant \[=0\] \[\therefore \] \[4{{(1+3k)}^{2}}=4\times 7(2k+3)\] \[\Rightarrow \] \[1+9{{k}^{2}}+6k=14k+21\] \[\Rightarrow \] \[9{{k}^{2}}-8k-20=0\] \[\Rightarrow \] \[(9k+10)\,(k-2)=0\] \[\Rightarrow \] \[k=2,\,\frac{-10}{9}\]


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