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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 a, b, g are the roots of the equation \[2{{x}^{3}}-3{{x}^{2}}+6x+1=0\], then \[{{\alpha }^{2}}+{{\beta }^{2}}+{{\gamma }^{2}}\] is equal to [Karnataka CET 2005]

    A) -\[\frac{15}{4}\]

    B) \[\frac{15}{4}\]

    C) \[\frac{9}{4}\]

    D) 4

    Correct Answer: A

    Solution :

    Given equation \[2{{x}^{3}}-3{{x}^{2}}+6x+1=0\] \[\alpha +\beta +\gamma =\frac{3}{2}\], \[\alpha \beta \gamma =\frac{-1}{2}\], \[\Sigma \alpha \beta =3\] \[({{\alpha }^{2}}+{{\beta }^{2}}+{{\gamma }^{2}})={{(\alpha +\beta +\gamma )}^{2}}-2(\Sigma \alpha \beta )\] = \[{{\left( \frac{3}{2} \right)}^{2}}-2.3\]= \[\frac{9}{4}-6=\frac{-15}{4}\].


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