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10th Class Mathematics Related to Competitive Exam Question Bank Algebra

  • question_answer
    The solution set of the equation \[{{x}^{\frac{2}{3}}}+{{x}^{\frac{1}{3}}}=2\] is

    A)  \[\{-8,1\}\]                          

    B)         {8,1}

    C) \[\text{ }\!\!\{\!\!\text{ 1},-\text{1}\}\]               

    D)         \[\{2,-2\}\]

    Correct Answer: A

    Solution :

     Let         \[{{x}^{\frac{1}{3}}}=y\] Then      \[{{y}^{2}}+y-2=0\] or            \[(y+2)(y-1)=0\] or            \[y=-2,\,\,1\] When    \[y=-2{{x}^{\frac{1}{3}}}=-2\] or            \[x=-8\] When    \[y=1{{x}^{\frac{1}{3}}}=1\] or            \[x=1\] Hence solution set is\[\{-8,\,\,1\}\].


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