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SSC Sample Paper Mock Test-11 SSC CGL Tear-II Paper-1

  • question_answer
    If \[x=3+\sqrt{8},\] then \[{{x}^{2}}+\frac{1}{{{x}^{2}}}\] equal  to

    A) 38

    B)  36      

    C)  34

    D)  30

    Correct Answer: C

    Solution :

    \[\because \]\[x=3+\sqrt{8}\] \[\therefore \]\[\frac{1}{x}=\frac{1}{3+\sqrt{8}}=\frac{3-\sqrt{8}}{(3+\sqrt{8})(3-\sqrt{8})}\] \[=\frac{3-\sqrt{8}}{9-8}=3-\sqrt{8}\] Now, \[{{x}^{2}}+\frac{1}{{{x}^{2}}}={{\left( x+\frac{1}{x} \right)}^{2}}-2\] \[={{(3+\sqrt{8}+3-\sqrt{8})}^{2}}-2\] \[=36-2=34\]


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