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JAMIA MILLIA ISLAMIA Jamia Millia Islamia Solved Paper-2007

  • question_answer
        If\[(1-p)\]is a root of quadratic equation \[{{x}^{2}}+px+(1-p)=0,\]then its roots are

    A)  0, 1                                       

    B)  \[-1,1\]

    C)  0,-1                                      

    D)  \[-1,2\]

    Correct Answer: C

    Solution :

                     Since\[(1-p)\]is the root of quadratic equation \[{{x}^{2}}+px+(1-p)=0\]                        ...(i) So,\[(1-p)\]satisfied the above equation. \[\therefore \]  \[{{(1-p)}^{2}}+p(1-p)+(1-p)=0\]                 \[(1-p)[1-p+p+1]=0\] \[\Rightarrow \]                               \[(1-p)(2)=0\] \[\Rightarrow \]                                               \[p=1\] On putting this value of p in Eq. (i) \[{{x}^{2}}+x=0\] \[\Rightarrow \]               \[x(x+1)=0\] \[\Rightarrow \]               \[x=0,-1\]


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