Manipal Engineering Manipal Engineering Solved Paper-2011

  • question_answer
    The square root of\[\sqrt{50}+\sqrt{48}\]is

    A) \[{{2}^{1/4}}(3+\sqrt{2})\]                          

    B) \[{{2}^{1/4}}(\sqrt{3}+\sqrt{2})\]

    C) \[{{2}^{1/4}}(2+\sqrt{2})\]          

    D)        \[{{2}^{1/4}}(\sqrt{3}+2)\]

    Correct Answer: B

    Solution :

    Now,\[\sqrt{50}+\sqrt{48}=5\sqrt{2}+4\sqrt{3}\]                 \[=\sqrt{2}[5+2\cdot \sqrt{2}\cdot \sqrt{3}]\]                 \[=\sqrt{2}{{(\sqrt{3}+\sqrt{2})}^{2}}\] \[\therefore \]  \[\sqrt{\sqrt{50}+\sqrt{48}}={{2}^{1/4}}(\sqrt{3}+\sqrt{2})\]


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