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JEE Main & Advanced Mathematics Complex Numbers and Quadratic Equations Question Bank Condition for common roots, Quadratic expressions and Position of roots

  • question_answer
    The maximum possible number of real roots of equation \[{{x}^{5}}-6{{x}^{2}}-4x+5=0\] is [EAMCET 2002]

    A) 0

    B) 3

    C) 4

    D) 5

    Correct Answer: B

    Solution :

    Let \[f(x)={{x}^{5}}-6{{x}^{2}}-4x+5=0\] Then the number of change of sign in \[f(x)\] is 2 therefore \[f(x)\] can have at most two positive real roots. Now, \[f(-x)=-{{x}^{5}}-6{{x}^{4}}+4x+5=0\] Then the number of change of sign is 1. Hence \[f(x)\]can have at most one negative real root. So that total possible number of real roots is 3.


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