Results 1 to 3 of 3

Thread: surds

  1. #1
    Newbie
    Joined
    Jun 2009
    Posts
    3

    surds

    I am having difficulty solving these surds problems

    $\displaystyle (5+2 \sqrt{7})^2$

    and

    $\displaystyle \sqrt{8}+\sqrt{18}-\sqrt{50}$
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor alexmahone's Avatar
    Joined
    Oct 2008
    Posts
    1,114
    Thanks
    7
    $\displaystyle (5+2\sqrt7)^2=25+28+20\sqrt{7}$
    $\displaystyle =53+20\sqrt{7}$

    $\displaystyle \sqrt{8}+\sqrt{18}-\sqrt{50}=2\sqrt{2}+3\sqrt{2}-5\sqrt{2}=0$
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor
    Prove It's Avatar
    Joined
    Aug 2008
    Posts
    12,880
    Thanks
    1946
    Quote Originally Posted by b3nj1d View Post
    I am having difficulty solving these surds problems

    $\displaystyle (5+2 \sqrt{7})^2$

    and

    $\displaystyle \sqrt{8}+\sqrt{18}-\sqrt{50}$
    Since we haven't learnt FOIL yet, I understand why you're having trouble with the first one.

    If you have two binomials like $\displaystyle (a + b)(c + d)$, they are expanded by multiplying the Firsts, then the Outers, then the Inners, then the Lasts.

    So $\displaystyle (a + b)(c + d) = ac + ad + bc + bd$.


    So for your first one, you have

    $\displaystyle (5 + 2\sqrt{7})^2 = (5 + 2\sqrt{7})(5 + 2\sqrt{7})$

    Multiplying the firsts gives $\displaystyle 5 \times 5 = 25$.

    Multiplying the outers gives $\displaystyle 5 \times 2\sqrt{7} = 10\sqrt{7}$

    Multiplying the inners gives $\displaystyle 5 \times 2\sqrt{7} = 10\sqrt{7}$

    Multiplying the lasts gives $\displaystyle 2\sqrt{7}\times 2\sqrt{7} = 4 \times 7 = 28$.


    So $\displaystyle (5 + 2\sqrt{7})^2 = 25 + 10\sqrt{7} + 10\sqrt{7} + 28$

    $\displaystyle = 53 + 20\sqrt{7}$.



    For the second, you need to simplify all of the surds first. Then hopefully you'll have "like surds" that can be collected.

    $\displaystyle \sqrt{8} = \sqrt{4 \times 2} = \sqrt{4} \times \sqrt{2} = 2\sqrt{2}$.


    $\displaystyle \sqrt{18} = \sqrt{9 \times 2} = \sqrt{9}\times\sqrt{2} = 3\sqrt{2}$.


    $\displaystyle \sqrt{50} = \sqrt{25 \times 2} = \sqrt{25}\times\sqrt{2} = 5\sqrt{2}$.



    So $\displaystyle \sqrt{8} + \sqrt{18} - \sqrt{50} = 2\sqrt{2} + 3\sqrt{2} - 5\sqrt{2}$

    $\displaystyle = 5\sqrt{2} - 5\sqrt{2}$

    $\displaystyle = 0\sqrt{2}$

    $\displaystyle = 0$.



    Chookas for the exam tomorrow.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. [SOLVED] Help with Surds
    Posted in the Algebra Forum
    Replies: 8
    Last Post: Jan 5th 2012, 03:08 PM
  2. Something to do with surds?
    Posted in the Algebra Forum
    Replies: 4
    Last Post: Jul 11th 2011, 10:33 AM
  3. surds
    Posted in the LaTeX Help Forum
    Replies: 3
    Last Post: Jun 4th 2009, 04:55 PM
  4. surds
    Posted in the Algebra Forum
    Replies: 1
    Last Post: May 29th 2009, 04:40 AM
  5. Surds
    Posted in the Math Topics Forum
    Replies: 1
    Last Post: Mar 1st 2009, 10:44 PM

Search Tags


/mathhelpforum @mathhelpforum