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JEE Main & Advanced Mathematics Complex Numbers and Quadratic Equations Question Bank Solution of quadratic inequations and Miscellaneous equations

  • question_answer
    If \[\alpha ,\,\beta ,\,\gamma \] are the roots of the equation \[{{x}^{3}}+4x+1=0,\] then \[{{(\alpha +\beta )}^{-1}}+{{(\beta +\gamma )}^{-1}}+{{(\gamma +\alpha )}^{-1}}=\] [EAMCET 2003]

    A) 2

    B) 3

    C) 4

    D) 5

    Correct Answer: C

    Solution :

    Ifa, b, g  are the roots of the equation.      \[{{x}^{3}}-p{{x}^{2}}+qx-r=0\] \[\therefore {{(\alpha +\beta )}^{-1}}+{{(\beta +\gamma )}^{-1}}+{{(\gamma +\alpha )}^{-1}}=\frac{{{p}^{2}}+q}{pq-r}\] Given, \[p=0,\,\,q=4,\,\,\,r=-1\] Þ \[\frac{{{p}^{2}}+q}{pq-r}=\frac{0+4}{0+1}=4\].


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