Results 1 to 5 of 5

Thread: prove

  1. #1
    Member
    Joined
    Nov 2006
    Posts
    152

    prove

    without a calculator

    prove
    $\displaystyle \sqrt{2+\sqrt{3}}+\sqrt{4-\sqrt{7}}=\sqrt{5+\sqrt{21}}$
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Bar0n janvdl's Avatar
    Joined
    Apr 2007
    From
    Meh
    Posts
    1,630
    Thanks
    6
    Quote Originally Posted by perash View Post
    without a calculator

    prove
    $\displaystyle \sqrt{2+\sqrt{3}}+\sqrt{4-\sqrt{7}}=\sqrt{5+\sqrt{21}}$
    You already posted this here
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Super Member wingless's Avatar
    Joined
    Dec 2007
    From
    Istanbul
    Posts
    585
    Quote Originally Posted by janvdl View Post
    You already posted this here
    ?

    -----

    $\displaystyle \sqrt{5+\sqrt{21}}=\sqrt{2+\sqrt{3}}+\sqrt{4-\sqrt{7}}$

    $\displaystyle 5+\sqrt{21}=(\sqrt{2+\sqrt{3}}+\sqrt{4-\sqrt{7}})^2$

    $\displaystyle 5+\sqrt{21}=2+\sqrt{3} + 4-\sqrt{7} + 2\sqrt{8 - 2\sqrt{7} + 4\sqrt{3} -\sqrt{21}}$

    $\displaystyle 2\sqrt{8 - 2\sqrt{7} + 4\sqrt{3} -\sqrt{21}} = \sqrt{21} - \sqrt{3} + \sqrt{7} - 1$

    $\displaystyle 8 - 2\sqrt{7} + 4\sqrt{3} -\sqrt{21} = (\frac{\sqrt{21} - \sqrt{3} + \sqrt{7} - 1}{2})^2$

    ............ (a few long but easy calculations here..)

    $\displaystyle 8 - 2\sqrt{7} + 4\sqrt{3} -\sqrt{21} = 8 - 2\sqrt{7} + 4\sqrt{3} -\sqrt{21} \text{ } \blacksquare$

    If you need help with the calculations I didn't show, just tell me
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Bar0n janvdl's Avatar
    Joined
    Apr 2007
    From
    Meh
    Posts
    1,630
    Thanks
    6
    Quote Originally Posted by wingless View Post
    ?

    -----
    He edited the post afterwards.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Math Engineering Student
    Krizalid's Avatar
    Joined
    Mar 2007
    From
    Santiago, Chile
    Posts
    3,656
    Thanks
    14
    Quote Originally Posted by perash View Post
    without a calculator

    prove
    $\displaystyle \sqrt{2+\sqrt{3}}+\sqrt{4-\sqrt{7}}=\sqrt{5+\sqrt{21}}$
    $\displaystyle \sqrt {2 + \sqrt 3 } = \frac{{\sqrt {4 + 2\sqrt 3 } }}
    {{\sqrt 2 }} = \frac{{\sqrt {\left( {1 + \sqrt 3 } \right)^2 } }}
    {{\sqrt 2 }} = \frac{{1 + \sqrt 3 }}
    {{\sqrt 2 }}.$

    In the same way $\displaystyle \sqrt {4 - \sqrt 7 } = \frac{{\sqrt 7 - 1}}
    {{\sqrt 2 }}.$

    Finally $\displaystyle \sqrt {2 + \sqrt 3 } + \sqrt {4 - \sqrt 7 } = \frac{{\sqrt 3 + \sqrt 7 }}
    {{\sqrt 2 }} = \frac{{\sqrt {10 + 2\sqrt {21} } }}
    {{\sqrt 2 }} = \sqrt {5 + \sqrt {21} } \quad\blacksquare$
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Prove a/b and a/c then a/ (3b-7c)
    Posted in the Discrete Math Forum
    Replies: 2
    Last Post: Mar 23rd 2010, 05:20 PM
  2. prove,,,
    Posted in the Number Theory Forum
    Replies: 1
    Last Post: Mar 1st 2010, 09:02 AM
  3. Prove |w + z| <= |w| +|z|
    Posted in the Algebra Forum
    Replies: 3
    Last Post: Feb 28th 2010, 05:44 AM
  4. Replies: 2
    Last Post: Aug 28th 2009, 02:59 AM
  5. How to prove that n^2 + n + 2 is even??
    Posted in the Algebra Forum
    Replies: 3
    Last Post: Nov 30th 2008, 01:24 PM

Search Tags


/mathhelpforum @mathhelpforum