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CET Karnataka Engineering CET - Karnataka Engineering Solved Paper-2008

  • question_answer
    The real root of the equation \[{{x}^{3}}-6x+9=0\] is

    A)  \[-6\]                                   

    B)  \[-9\]

    C)  \[6\]                                    

    D)  \[-3\]

    Correct Answer: D

    Solution :

    Given,  \[{{x}^{3}}-6x+9=0\] \[\Rightarrow \]               \[(x+3)({{x}^{2}}-3x+3)=0\] \[\Rightarrow \]               \[x=-3\]  or \[{{x}^{2}}-3x+3=0\] Now, Discriminant,   \[D=\sqrt{9-4\times 3}\] \[=\sqrt{-3}\] imaginary Hence, real roots of the given equation is                                 \[x=-3\]


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