Results 1 to 3 of 3
Like Tree2Thanks
  • 2 Post By Debsta

Thread: Solving radicals in the denominator AND flipping the sign(?)

  1. #1
    Newbie
    Joined
    Feb 2019
    From
    Illinois
    Posts
    2

    Question Solving radicals in the denominator AND flipping the sign(?)

    Hello Math Community,

    I am a 34 year old who hasn't practiced math in a while. I am usually able to help my kiddo with assignments, but I am simply stumped on this problem. I have tried several different ways to research why it's necessary to flip the sign when solving radicals in the denominator, but I am struggling to find answers: I have, 1.) looked in the algebra book, 2.) looked at my kiddo's notes 3.) googled it.

    I am not looking for a solution to this problem. I am looking to find WHY its necessary to flip the sign to be able to remove the radical from the denominator. Can you help me?

    Solve:

    (5) / (-5 - 3 * sqrt(3) )



    Solution as determined by my kiddo's notes & photomath's app:

    The step that confuses me is the bolded portion below. Why is it appropriate to multiply by a positive 3*sqrt(3) and not the negative 3*sqrt(3) like the denominator in the original expression above?

    [ (5) * (-5 + 3 * sqrt(3) ) ] / [ (-5 - 3 * sqrt(3) ) * (-5 + 3 * sqrt(3) ) ]



    Secondly, is there a resource that someone can recommend for me to use on how to solve problems like this? I am a fan of Khan Academy, but I am not sure why topic to search for.


    Thank you for your help community.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Joined
    Oct 2009
    From
    Brisbane
    Posts
    1,132
    Thanks
    376

    Re: Solving radicals in the denominator AND flipping the sign(?)

    It's all got to do with the fact that $\displaystyle (a+b)(a-b) = a^2 -ab +ab -b^2 =a^2 - b^2$ which is commonly called the difference of two squares.

    You can see how the middle two terms in the second expression will always cancel each other out. Squaring a square root will get rid of the square root sign.

    In your case: $\displaystyle (-5 - 3\sqrt{3})(-5+3\sqrt{3}) = (-5)^2 -15\sqrt{3} +15\sqrt{3} -(3\sqrt{3})^2 = 25 -27 =-2$

    See how the middle terms cancel out. You are left with a rational number (hence the term "rationalise the denominator").

    Try multiplying $\displaystyle (-5 -3\sqrt{3})(-5-3\sqrt{3})$ , (that is, without changing the sign to +).

    You will see that the middle terms don't cancel and you will not have a rational result.
    Last edited by Debsta; Feb 8th 2019 at 02:59 PM.
    Thanks from Emyk84 and HallsofIvy
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie
    Joined
    Feb 2019
    From
    Illinois
    Posts
    2

    Re: Solving radicals in the denominator AND flipping the sign(?)

    Thank you very much! That made COMPLETE sense. And I absolutely understand everything you said.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 1
    Last Post: Mar 1st 2013, 04:26 AM
  2. Solving for x in radicals
    Posted in the Algebra Forum
    Replies: 12
    Last Post: Oct 20th 2011, 07:05 AM
  3. Replies: 9
    Last Post: Nov 13th 2010, 02:27 AM
  4. [SOLVED] Rationalizing radicals in the denominator
    Posted in the Algebra Forum
    Replies: 20
    Last Post: Oct 9th 2010, 11:39 AM
  5. flipping the inequality sign
    Posted in the Algebra Forum
    Replies: 3
    Last Post: Mar 16th 2009, 08:29 PM

/mathhelpforum @mathhelpforum