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Math Help - Urgent help required !

  1. #1
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    Exclamation Urgent help required !

    Dear Sir ,
    I felt difficulty in solving the following problems .So please help me to solve these questions . I will be very thankful to you .\

    The question is in the attachment .
    Attached Thumbnails Attached Thumbnails Urgent help required !-doubt.jpg  
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  2. #2
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    Re: Urgent help required !

    Hello, rsaravanan!

    Here is the first one . . .


    \text{1) If }\,a \:=\:7-4\sqrt{3},\;\text{ find: }\,\sqrt{a} + \frac{1}{\sqrt{a}}

    \text{Note that: }\:a \;=\;7 - 4\sqrt{3} \;=\; (2-\sqrt{3})^2

    \text{Hence: }\:\sqrt{a} \;=\;\sqrt{(2-\sqrt{3})^2} \;=\;2-\sqrt{3}


    \text{Then: }\:\sqrt{a} + \frac{1}{\sqrt{a}} \;=\;\left(2-\sqrt{3}\right) + \frac{1}{2-\sqrt{3}}


    \text{Rationalize: }\:\left(2-\sqrt{3}\right)\;+\; \frac{1}{2-\sqrt{3}}\!\cdot\!{\color{blue}\frac{2+\sqrt{3}}{2  +\sqrt{3}}} \;\;=\;\;\left(2-\sqrt{3}\right)\;+\;\frac{2+\sqrt{3}}{4-3}

    . . . . . . . . . . =\;\;\left(2 - \sqrt{3}\right) + \left(2 + \sqrt{3}\right) \;\;=\;\;\boxed{4}
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  3. #3
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    Re: Urgent help required !

    Thank you very much sir
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  4. #4
    MHF Contributor MarkFL's Avatar
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    Re: Urgent help required !

    In the second problem, all three terms should be cubed. Also, just a heads up, I solved it on another forum on which these questions were posted.
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