Results 1 to 7 of 7

Thread: Radicals

  1. #1
    Member
    Joined
    Jan 2010
    Posts
    142

    Unhappy Radicals

    Can anyone solve this and explain it to me please. thank you!~!

    Sqrt of 3 / 1 - Ceubroot of 2 + (Sqrt of 3)(Cuberoot of 2 )

    Follow Math Help Forum on Facebook and Google+

  2. #2
    Super Member
    Joined
    Dec 2009
    From
    1111
    Posts
    872
    Thanks
    4
    Dear Anemori,

    Is this what you mean, $\displaystyle \frac{\sqrt{3}}{1-\sqrt[3]{2}}+(\sqrt{3}\times\sqrt[3]{2})$
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Member
    Joined
    Jan 2010
    Posts
    142
    Quote Originally Posted by Sudharaka View Post
    Dear Anemori,

    Is this what you mean, $\displaystyle \frac{\sqrt{3}}{1-\sqrt[3]{2}}+(\sqrt{3}\times\sqrt[3]{2})$



    Yes sir, that's what i meant.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor

    Joined
    Apr 2005
    Posts
    19,781
    Thanks
    3030
    Quote Originally Posted by Sudharaka View Post
    Dear Anemori,

    Is this what you mean, $\displaystyle \frac{\sqrt{3}}{1-\sqrt[3]{2}}+(\sqrt{3}\times\sqrt[3]{2})$
    Quote Originally Posted by Anemori View Post
    Yes sir, that's what i meant.
    Okay, then, what do you want to do with it? That is an "expression", not a "problem". Are you trying to add the two? To do that you will have to get fractions with the same denominator by multiplying $\displaystyle \sqrt{3}\sqrt[3]{2}$ by the unit fraction $\displaystyle \frac{1-\sqrt[3]{2}}{1-\sqrt[3]{2}}$.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Member
    Joined
    Jan 2010
    Posts
    142
    Quote Originally Posted by HallsofIvy View Post
    Okay, then, what do you want to do with it? That is an "expression", not a "problem". Are you trying to add the two? To do that you will have to get fractions with the same denominator by multiplying $\displaystyle \sqrt{3}\sqrt[3]{2}$ by the unit fraction $\displaystyle \frac{1-\sqrt[3]{2}}{1-\sqrt[3]{2}}$.

    Simplified it. can you explain how do you do it? i understand you cannot multiply or divide both radicals with different index.
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Super Member
    Joined
    Dec 2009
    From
    1111
    Posts
    872
    Thanks
    4
    Dear Anemori,

    $\displaystyle \frac{\sqrt{3}}{1-\sqrt[3]{2}}+(\sqrt{3}\times\sqrt[3]{2})$

    $\displaystyle \frac{\sqrt{3}+(1-\sqrt[3]{2})\sqrt{3}\sqrt[3]{2}}{1-\sqrt[3]{2}}$

    $\displaystyle \frac{\sqrt{3}(1-\sqrt[3]{2}-\sqrt{3}(\sqrt[3]{2})^2)}{1-\sqrt[3]{2}}$

    $\displaystyle \frac{\sqrt{3}(1-\sqrt[3]{2}-2^{\frac{2}{3}})}{1-\sqrt[3]{2}}$

    Hope this solves the questions you have about radical expressions.
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Super Member

    Joined
    May 2006
    From
    Lexington, MA (USA)
    Posts
    12,028
    Thanks
    848
    Hello, Anemori!

    Simplify: .$\displaystyle \frac{\sqrt{3}}{1 - \sqrt[3]{2}} + \sqrt{3}\cdot\sqrt[3]{2} $

    Factor out $\displaystyle \sqrt{3}\!:\;\;\sqrt{3}\,\bigg[\frac{1}{1-\sqrt[3]{2}} + \sqrt[3]{2}\bigg]$


    Multiply the fraction by $\displaystyle \frac{1 + \sqrt[3]{2} + \sqrt[3]{4}}{1 + \sqrt[3]{2} + \sqrt[3]{4}} $

    . . $\displaystyle \frac{1}{1-\sqrt[3]{2}}\cdot\frac{1 + \sqrt[3]{2} + \sqrt[3]{4}}{1+\sqrt[3]{2} + \sqrt[3]{4}} \;=\; \frac{1+\sqrt[3]{2} + \sqrt[3]{4}}{1-2} \;=\;-\left(1 + \sqrt[3]{2} + \sqrt[3]{4}\right)$ .$\displaystyle =\;-1-\sqrt[3]{2} - \sqrt[3]{4}$


    The problem becomes: . $\displaystyle \sqrt{3}\bigg[-1 - \sqrt[3]{2} - \sqrt[3]{4} + \sqrt[3]{2}\bigg] \;=\;\sqrt{3}\bigg[-1 - \sqrt[3]{4}\bigg]$


    Therefore: .$\displaystyle -\sqrt{3}\left(1 + \sqrt[3]{4}\right) $

    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. help with radicals
    Posted in the Algebra Forum
    Replies: 4
    Last Post: May 4th 2010, 07:49 AM
  2. Help with radicals :(
    Posted in the Algebra Forum
    Replies: 2
    Last Post: Dec 2nd 2009, 08:20 PM
  3. Radicals
    Posted in the Algebra Forum
    Replies: 5
    Last Post: Apr 23rd 2009, 05:19 AM
  4. Radicals
    Posted in the Algebra Forum
    Replies: 4
    Last Post: Jan 22nd 2008, 02:40 AM
  5. Radicals
    Posted in the Algebra Forum
    Replies: 1
    Last Post: Oct 17th 2007, 08:24 AM

Search Tags


/mathhelpforum @mathhelpforum