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RAJASTHAN ­ PET Rajasthan PET Solved Paper-2010

  • question_answer
    The value of kfor which the roots of the equation\[{{x}^{2}}+2(k+1)x+{{k}^{2}}=0\]are equal to

    A)  \[-1\]             

    B)  \[-\frac{1}{2}\]

    C)  \[1\]              

    D)  \[\frac{1}{2}\]

    Correct Answer: B

    Solution :

     Since roots of the equation \[{{x}^{2}}+2(k+1)x+{{k}^{2}}=0\] are equal. \[\therefore \] \[{{B}^{2}}=4AC\] \[\Rightarrow \] \[4{{(k+1)}^{2}}=4{{k}^{2}}\] \[\Rightarrow \] \[4{{k}^{2}}+4+8k=4{{k}^{2}}\] \[\Rightarrow \] \[k=-\frac{1}{2}\]


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