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

  • question_answer
    One root of the following given equation \[2{{x}^{5}}-14{{x}^{4}}+31{{x}^{3}}-64{{x}^{2}}+19x+130=0\] is  [MP PET 1985]

    A) 1

    B) 3

    C) 5

    D) 7

    Correct Answer: C

    Solution :

    Putting \[x=5\] \[2{{(5)}^{5}}-14{{(5)}^{4}}+31{{(5)}^{3}}-64{{(5)}^{2}}+19(5)+130=0\] Hence \[x=5\] satisfies the given equation. Thus 5 is a root of the equation.


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