NTSE Sample Paper NTSE SAT Practice Test-12

  • question_answer
    What is the square root of \[23+4\sqrt{10}-10\sqrt{2}-8\sqrt{5}?\]

    A) \[\sqrt{5}+\sqrt{10}-2\sqrt{2}\]

    B)        \[\sqrt{5}-\sqrt{10}+2\sqrt{5}\]

    C) \[\sqrt{5}+2\sqrt{2}-\sqrt{10}\]

    D)        \[2\sqrt{2}-\sqrt{5}+\sqrt{10}\]

    Correct Answer: C

    Solution :

    [c] \[{{(a+b-c)}^{2}}={{a}^{2}}+{{b}^{2}}+{{c}^{2}}+2ab-2bc-2ac\] we have, \[{{(\sqrt{5}+2\sqrt{2}-\sqrt{10})}^{2}}=5+8+10\] \[+4\sqrt{10}-8\sqrt{5}-10\sqrt{2}\] or \[\sqrt{5}+2\sqrt{2}-\sqrt{10}\] \[=\sqrt{23+4\sqrt{10}-10\sqrt{2}-8\sqrt{5}}\] LCM \[={{a}^{3}}+{{a}^{2}}-a-1=(a+1)({{a}^{2}}-1)\] HCF \[\times \] LCM = Product of numbers \[\therefore \text{ (}a+1)\times \left( a+1 \right)\text{ }\left( {{a}^{2}}-1 \right)=\left( {{a}^{2}}-1 \right)\text{ }\times \]Second number or Second number \[=\text{ }{{\left( a\text{ }+\text{ 1} \right)}^{2}}\]


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