Results 1 to 8 of 8
Like Tree3Thanks
  • 1 Post By Shakarri
  • 2 Post By agentmulder

Math Help - Simplifiy this radical

  1. #1
    Member
    Joined
    Oct 2011
    Posts
    170
    Thanks
    3

    Simplifiy this radical

    \sqrt{\frac{4a^3}{27b^3}}

    =\sqrt{\frac{2\cdot 2\cdot a\cdot a\cdot a}{9\cdot 3\cdot b\cdot b\cdot b}} and I can take the square root of 9.

    =6ab\sqrt{\frac{a}{b}} is this correct?

    The book I'm working from doesn't have the answers so I'm cross checking my work with Wolfram but Wolfram has a habbit of providing the same answer but in a totally different format. If I'm wrong please explain why thank you

    This is the solution from Wolfram \frac{2\sqrt{\frac{a^3}{b^3}}}{3\sqrt{3}}

    **EDIT** as I relook over the problem, maybe the answer might be \frac{2a}{3b}\sqrt{\frac{a}{3b}} ?
    Last edited by uperkurk; August 12th 2013 at 04:48 AM.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Super Member
    Joined
    Oct 2012
    From
    Ireland
    Posts
    591
    Thanks
    159

    Re: Simplifiy this radical

    Quote Originally Posted by uperkurk View Post
    \sqrt{\frac{4a^3}{27b^3}}

    =\sqrt{\frac{2\cdot 2\cdot a\cdot a\cdot a}{9\cdot 3\cdot b\cdot b\cdot b}} and I can take the square root of 9.
    You are correct up to here then you get the fractions confused

    Collect pairs of the same terms together so you can get the square root easily
    =\sqrt{\frac{2^2\cdot a^2 \cdot a}{3^2\cdot 3\cdot b^2\cdot b}}

    Then split the square root into those terms which are squared and those which aren't
    =\sqrt{\frac{2^2\cdot a^2}{3^2\cdot b^2}}\sqrt{\frac{a}{b\cdot 3}}

    =\frac{2\cdot a}{3\cdot b}\sqrt{\frac{a}{b\cdot 3}}
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Member
    Joined
    Oct 2011
    Posts
    170
    Thanks
    3

    Re: Simplifiy this radical

    Quote Originally Posted by Shakarri View Post

    =\frac{2\cdot a}{3\cdot b}\sqrt{\frac{a}{b\cdot 3}}
    Thanks for your help, but is this not the exact same thing I have?
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Super Member
    Joined
    Oct 2012
    From
    Ireland
    Posts
    591
    Thanks
    159

    Re: Simplifiy this radical

    Ah yes it is, as I was posting you edited your post saying that you had looked over the problem again. Wolfram Alpha just expresses the answer differently.
    Thanks from uperkurk
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Member
    Joined
    Oct 2011
    Posts
    170
    Thanks
    3

    Re: Simplifiy this radical

    Quote Originally Posted by Shakarri View Post
    Ah yes it is, as I was posting you edited your post saying that you had looked over the problem again. Wolfram Alpha just expresses the answer differently.
    ok so I'm looking at this one, hopefully you can point me in the right direction.

    \sqrt[4]{\frac{2a^8}{b^2c^3}}

    I'll say I need to do something like this \sqrt[4]{\frac{2a^8}{b^2c^3}}\times \sqrt[4]{\frac{27b}{27b}} = \sqrt[4]{\frac{54a^8b}{3^3b^3c^3}} so now I have the denominators all the same, now what? I could take the cube root of most things?

    \sqrt[4]{\frac{54a^8b}{3^3b^3c^3}} = \frac{\sqrt[4]{6\cdot 9\cdot a^8\cdot b}}{\sqrt{3bc}} I'm lost here...

    **EDIT** nah that surely isn't right, the 4th root is making it confusing
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Junior Member
    Joined
    Jan 2010
    Posts
    52
    Thanks
    17

    Re: Simplifiy this radical

    Quote Originally Posted by uperkurk View Post
    ok so I'm looking at this one, hopefully you can point me in the right direction.

    \sqrt[4]{\frac{2a^8}{b^2c^3}}

    I'll say I need to do something like this \sqrt[4]{\frac{2a^8}{b^2c^3}}\times \sqrt[4]{\frac{27b}{27b}} = \sqrt[4]{\frac{54a^8b}{3^3b^3c^3}} so now I have the denominators all the same, now what? I could take the cube root of most things?

    \sqrt[4]{\frac{54a^8b}{3^3b^3c^3}} = \frac{\sqrt[4]{6\cdot 9\cdot a^8\cdot b}}{\sqrt{3bc}} I'm lost here...

    **EDIT** nah that surely isn't right, the 4th root is making it confusing
    It shouldn't as it is true that:
    \sqrt[R]{A} = {A^{\frac{1}{R}}}
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Super Member
    Joined
    Oct 2012
    From
    Ireland
    Posts
    591
    Thanks
    159

    Re: Simplifiy this radical

    I'll say I need to do something like this \sqrt[4]{\frac{2a^8}{b^2c^3}}\times \sqrt[4]{\frac{27b}{27b}}
    There should be no reason to multiply by \sqrt[4]{\frac{27b}{27b}} if you are going to simplify it immediately. You could simplify it before bringing them into 1 fraction
    \sqrt[4]{\frac{2a^8}{b^2c^3}}\times \sqrt[4]{\frac{27b}{27b}}=\sqrt[4]{\frac{2a^8}{b^2c^3}}\times 1

    Learn the rules of indices for taking roots. When you have \sqrt[4]{a^3} that is equal to (a^3)^\frac{1}{4} which in turn is equal to a^\frac{3}{4}
    Follow Math Help Forum on Facebook and Google+

  8. #8
    Member agentmulder's Avatar
    Joined
    Aug 2011
    From
    42nd parallel, North America
    Posts
    105
    Thanks
    33

    Re: Simplifiy this radical

    Quote Originally Posted by uperkurk View Post
    ok so I'm looking at this one, hopefully you can point me in the right direction.

    \sqrt[4]{\frac{2a^8}{b^2c^3}}

    I'll say I need to do something like this \sqrt[4]{\frac{2a^8}{b^2c^3}}\times \sqrt[4]{\frac{27b}{27b}} = \sqrt[4]{\frac{54a^8b}{3^3b^3c^3}} so now I have the denominators all the same, now what? I could take the cube root of most things?

    \sqrt[4]{\frac{54a^8b}{3^3b^3c^3}} = \frac{\sqrt[4]{6\cdot 9\cdot a^8\cdot b}}{\sqrt{3bc}} I'm lost here...

    **EDIT** nah that surely isn't right, the 4th root is making it confusing
    You want to 'rationalize the denominator' , that means the FACTORS of the denominator should become perfect 4th powers.

     \sqrt[4]{ \frac{2a^8}{b^2c^3} \cdot \frac{b^2c}{b^2c}} \ = \ \frac{ \sqrt[4]{2a^8b^2c}}{ \sqrt[4]{b^4c^4}} \ = \ \frac{a^2 \sqrt[4]{2b^2c}}{bc}

    Thanks from MarkFL and uperkurk
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. [SOLVED] Ring isomorphism maps a Jacobson radical into a Jacobson radical?
    Posted in the Advanced Algebra Forum
    Replies: 4
    Last Post: June 21st 2010, 10:45 AM
  2. Radical-Radical equation
    Posted in the Algebra Forum
    Replies: 6
    Last Post: June 27th 2009, 03:38 AM
  3. f(x)=x^2/3 (x^2-4)...simplifiy then get f'(x)
    Posted in the Calculus Forum
    Replies: 1
    Last Post: March 26th 2009, 01:39 PM
  4. radical
    Posted in the Algebra Forum
    Replies: 2
    Last Post: July 22nd 2007, 02:02 AM
  5. Radical inside a radical?
    Posted in the Algebra Forum
    Replies: 3
    Last Post: March 14th 2007, 09:59 PM

Search Tags


/mathhelpforum @mathhelpforum