Results 1 to 6 of 6

Math Help - Surds

  1. #1
    Member
    Joined
    Jul 2011
    Posts
    140

    Surds

    I'm having a bit of trouble with this one. Can anyone help me out?

    Many thanks.


    Q. Given that (a+\sqrt{3})(b-\sqrt{3})=7+3\sqrt{3} & that a & b are positive integers, find the value of a & b.

    Attempt: ab-a\sqrt{3}+b\sqrt{3}=7+3\sqrt{3} => ab-a\sqrt{3}=3\sqrt{3}-b\sqrt{3}+10 => a(b-\sqrt{3})=3\sqrt{3}-b\sqrt{3}+10 => a=\frac{3\sqrt{3}-b\sqrt{3}+10}{b-\sqrt{3}} => a=\frac{\sqrt{3}(3-b)+10}{b-\sqrt{3}} => a=\frac{\sqrt{3}(3-b)+10}{b-\sqrt{3}}.\frac{b+\sqrt{3}}{b+\sqrt{3}} => a=\frac{3b\sqrt{3}-b^2\sqrt{3}+10b+9-3b+10\sqrt{3}}{b^2-3}...

    Ans.: (From text book): a = 2, b = 5
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Prove It's Avatar
    Joined
    Aug 2008
    Posts
    11,408
    Thanks
    1294

    Re: Surds

    Quote Originally Posted by GrigOrig99 View Post
    I'm having a bit of trouble with this one. Can anyone help me out?

    Many thanks.


    Q. Given that (a+\sqrt{3})(b-\sqrt{3})=7+3\sqrt{3} & that a & b are positive integers, find the value of a & b.

    Attempt: ab-a\sqrt{3}+b\sqrt{3}=7+3\sqrt{3} => ab-a\sqrt{3}=3\sqrt{3}-b\sqrt{3}+10 => a(b-\sqrt{3})=3\sqrt{3}-b\sqrt{3}+10 => a=\frac{3\sqrt{3}-b\sqrt{3}+10}{b-\sqrt{3}} => a=\frac{\sqrt{3}(3-b)+10}{b-\sqrt{3}} => a=\frac{\sqrt{3}(3-b)+10}{b-\sqrt{3}}.\frac{b+\sqrt{3}}{b+\sqrt{3}} => a=\frac{3b\sqrt{3}-b^2\sqrt{3}+10b+9-3b+10\sqrt{3}}{b^2-3}...

    Ans.: (From text book): a = 2, b = 5
    \displaystyle \begin{align*} \left(a + \sqrt{3}\right)\left(b - \sqrt{3}\right) &= 7 + 3\sqrt{3} \\ ab - a\sqrt{3} + b\sqrt{3} - 3 &= 7 + 3\sqrt{3} \\ ab - 3 + \left(b - a\right)\sqrt{3} &= 7 + 3\sqrt{3} \\ ab - 3 = 7 \textrm{ and } b - a &= 3 \\ ab = 10 \textrm{ and } b - a &= 3 \end{align*}

    Now solve these equations simultaneously for a and b.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor

    Joined
    Aug 2006
    Posts
    18,605
    Thanks
    1574
    Awards
    1

    Re: Surds

    Quote Originally Posted by GrigOrig99 View Post
    Q. Given that (a+\sqrt{3})(b-\sqrt{3})=7+3\sqrt{3} & that a & b are positive integers, find the value of a & b. Ans.: (From text book): a = 2, b = 5
    (a+\sqrt{3})(b-\sqrt{3})=7+3\sqrt{3}
    (ab-3)+(-a+b)\sqrt3=7+3\sqrt{3}

    So ab-3=7~\&~-a+b=3.

    Solve for a~\&~b.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Newbie
    Joined
    Jun 2012
    From
    hk
    Posts
    13
    Thanks
    3

    Re: Surds

    From above, then we let b, -a are root of a quadric equation s.t. -ab=-10, b+(-a)=3
    ie, equation is x^2-3x-10=0
    (x+2)(x-5)=0

    then (a=2,b=5) or (a=5,b=2)
    by putting to original equation, answer is found
    Follow Math Help Forum on Facebook and Google+

  5. #5
    MHF Contributor

    Joined
    Apr 2005
    Posts
    15,413
    Thanks
    1328

    Re: Surds

    Quote Originally Posted by GrigOrig99 View Post
    I'm having a bit of trouble with this one. Can anyone help me out?

    Many thanks.


    Q. Given that (a+\sqrt{3})(b-\sqrt{3})=7+3\sqrt{3} & that a & b are positive integers, find the value of a & b.

    Attempt: ab-a\sqrt{3}+b\sqrt{3}=7+3\sqrt{3}
    There's an error in your first line. You have "FOI" but not "L"! (-\sqrt{3})(\sqrt{3}= -3
    so this should be ab- a\sqrt{3}+ b\sqrt{3}- 3= 7+ 3\sqrt{3}
    You have ab- 3= 7 and b- a= 3

    => ab-a\sqrt{3}=3\sqrt{3}-b\sqrt{3}+10 => a(b-\sqrt{3})=3\sqrt{3}-b\sqrt{3}+10 => a=\frac{3\sqrt{3}-b\sqrt{3}+10}{b-\sqrt{3}} => a=\frac{\sqrt{3}(3-b)+10}{b-\sqrt{3}} => a=\frac{\sqrt{3}(3-b)+10}{b-\sqrt{3}}.\frac{b+\sqrt{3}}{b+\sqrt{3}} => a=\frac{3b\sqrt{3}-b^2\sqrt{3}+10b+9-3b+10\sqrt{3}}{b^2-3}...

    Ans.: (From text book): a = 2, b = 5
    a= -5, b= -2 is also a solution.
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Member
    Joined
    Jul 2011
    Posts
    140

    Re: Surds

    Great. Thanks guys.
    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: January 5th 2012, 03:08 PM
  2. Something to do with surds?
    Posted in the Algebra Forum
    Replies: 4
    Last Post: July 11th 2011, 10:33 AM
  3. surds
    Posted in the LaTeX Help Forum
    Replies: 3
    Last Post: June 4th 2009, 04:55 PM
  4. surds
    Posted in the Algebra Forum
    Replies: 2
    Last Post: June 4th 2009, 12:30 AM
  5. surds
    Posted in the Algebra Forum
    Replies: 1
    Last Post: May 29th 2009, 04:40 AM

Search Tags


/mathhelpforum @mathhelpforum