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JEE Main & Advanced Mathematics Complex Numbers and Quadratic Equations Question Bank De Moivre's theorem and Roots of unity

  • question_answer
     If \[\frac{1+\sqrt{3}\,i}{2}\] is a root of equation \[{{x}^{4}}-{{x}^{3}}+x-1=0\]  then its real roots are [EAMCET 2002]

    A) 1, 1

    B) - 1, - 1

    C) 1, - 1

    D) 1, 2

    Correct Answer: C

    Solution :

    \[{{x}^{4}}-{{x}^{3}}+x-1=0\] Þ \[{{x}^{3}}(x-1)+1(x-1)=0\] \[x-1=0\] or \[{{x}^{3}}+1=0\] Þ\[x=1,\,-1,\,\frac{1+\sqrt{3}i}{2},\,\,\frac{1-\sqrt{3}i}{2}\] so its real roots are 1 and - 1.


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