Results 1 to 11 of 11

Math Help - Simplifying roots

  1. #1
    Newbie
    Joined
    Apr 2010
    Posts
    13

    Simplifying roots

    Alright, I've got a final in College Algebra tomorrow but can't figure out how my teacher got this answer. Here was the original problem (as best I can type it):

    1 over the 5th root of 9

    I am told to rationalize the denominator. I got to this:

    5th root of 9 to the 4th power over 9

    This works as the answer, but it can simplify even further apparently. I can't figure out how it simplified further into:

    5th root of 27 over 3

    Help?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    skeeter's Avatar
    Joined
    Jun 2008
    From
    North Texas
    Posts
    11,690
    Thanks
    448
    Quote Originally Posted by Kieth89 View Post
    Alright, I've got a final in College Algebra tomorrow but can't figure out how my teacher got this answer. Here was the original problem (as best I can type it):

    1 over the 5th root of 9

    I am told to rationalize the denominator. I got to this:

    5th root of 9 to the 4th power over 9

    This works as the answer, but it can simplify even further apparently. I can't figure out how it simplified further into:

    5th root of 27 over 3

    Help?
    \displaystyle \frac{\sqrt[5]{9^4}}{9} = \frac{\sqrt[5]{3^8}}{9} = \frac{\sqrt[5]{3^5 \cdot 3^3}}{9} = \frac{3\sqrt[5]{3^3}}{9} = \frac{\sqrt[5]{3^3}}{3}
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie
    Joined
    Apr 2010
    Posts
    13
    How did you get the root sign to type
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor
    Joined
    Dec 2009
    Posts
    3,120
    Thanks
    1
    Quote Originally Posted by Kieth89 View Post
    Alright, I've got a final in College Algebra tomorrow but can't figure out how my teacher got this answer. Here was the original problem (as best I can type it):

    1 over the 5th root of 9

    I am told to rationalize the denominator. I got to this:

    5th root of 9 to the 4th power over 9

    This works as the answer, but it can simplify even further apparently. I can't figure out how it simplified further into:

    5th root of 27 over 3

    Help?

    \displaystyle\frac{1}{\sqrt[5]{9}}=\frac{1}{\sqrt[5]{3^2}}=\frac{1}{3^{\frac{2}{5}}}=\frac{3^{\frac{3}  {5}}}{3^{\frac{2}{5}}\;3^{\frac{3}{5}}}
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Newbie
    Joined
    Apr 2010
    Posts
    13
    Could you help me with this one, too?

    \sqrt{(3+x)^2-(3-x)^2}

    Alright, I know that I can't take the things out of the root until they are being multiplied, so what I did at first was just what the problem says:

    \sqrt{(3+x)(3+x)+(-3+x)(-3+x)} = \sqrt{9+6x+x^2+9-6x+x^2}<br />
 = \sqrt{2x^2+18} =
    \sqrt{2(X^2+9)} = ?
    I don't know if I messed up somewhere, or am not seeing the key, but I'm not getting any further than that. The correct answer is supposed to be 2\sqrt{3x}
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Super Member bigwave's Avatar
    Joined
    Nov 2009
    From
    honolulu
    Posts
    580

    another way

    by fractional exponents (probably not any easier tho)

     <br />
\frac{1}{9^{\frac{1}{5}}} \rightarrow \frac{1}{3^{\frac{2}{5}}}<br />

     <br />
\frac{1}{3^{\frac{2}{5}}}<br />
\times<br />
\frac<br />
{3^<br />
{\frac{3}{5}}<br />
}<br />
{3^<br />
{\frac{3}{5}}<br />
}<br />
\rightarrow<br />
\frac<br />
{3^{\frac{3}{5}}}<br />
{3^{\frac{5}{5}}}<br />
\rightarrow<br />
\frac<br />
{27^<br />
{\frac{1}{5}}<br />
}<br />
{3}<br />
    Follow Math Help Forum on Facebook and Google+

  7. #7
    MHF Contributor
    Prove It's Avatar
    Joined
    Aug 2008
    Posts
    11,548
    Thanks
    1418
    \displaystyle \sqrt{(3+x)^2 - (3 - x)^2} = \sqrt{9 + 6x + x^2 - (9 - 6x + x^2)}

    \displaystyle = \sqrt{9 + 6x + x^2 - 9 + 6x - x^2}

    \displaystyle = \sqrt{12x}

    \displaystyle = \sqrt{4\cdot 3x}

    \displaystyle =\sqrt{4}\cdot \sqrt{3x}

    \displaystyle = 2\sqrt{3x}
    Follow Math Help Forum on Facebook and Google+

  8. #8
    Newbie
    Joined
    Apr 2010
    Posts
    13
    Quote Originally Posted by Prove It View Post
    \displaystyle \sqrt{(3+x)^2 - (3 - x)^2} = \sqrt{9 + 6x + x^2 - (9 - 6x + x^2)}

    \displaystyle = \sqrt{9 + 6x + x^2 - 9 + 6x - x^2}

    \displaystyle = \sqrt{12x}

    \displaystyle = \sqrt{4\cdot 3x}

    \displaystyle =\sqrt{4}\cdot \sqrt{3x}

    \displaystyle = 2\sqrt{3x}

    So did distributing the minus first mess it all up for me?
    Follow Math Help Forum on Facebook and Google+

  9. #9
    Super Member bigwave's Avatar
    Joined
    Nov 2009
    From
    honolulu
    Posts
    580
     <br />
\sqrt{9 +6x+x^2-9+6x-x^2}<br />
\rightarrow<br />
\sqrt{12x}<br />
\rightarrow<br />
\sqrt{3}\sqrt{4}\sqrt{x}<br />
\rightarrow<br />
2\sqrt{3x}<br />
    Last edited by bigwave; December 14th 2010 at 05:39 PM. Reason: oh well, too late again
    Follow Math Help Forum on Facebook and Google+

  10. #10
    MHF Contributor
    skeeter's Avatar
    Joined
    Jun 2008
    From
    North Texas
    Posts
    11,690
    Thanks
    448
    Quote Originally Posted by Kieth89 View Post
    So did distributing the minus first mess it all up for me?
    order of operations, remember? exponents first, then multiply by -1 ...

    -(3-x)^2 = -(9 - 6x + x^2) = -9 + 6x - x^2
    Follow Math Help Forum on Facebook and Google+

  11. #11
    Newbie
    Joined
    Apr 2010
    Posts
    13
    Oh...it's always something simple..

    Thanks for the help
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. simplifying square roots
    Posted in the Algebra Forum
    Replies: 2
    Last Post: March 15th 2011, 03:35 AM
  2. Simplifying square roots?
    Posted in the Algebra Forum
    Replies: 2
    Last Post: May 11th 2010, 10:06 AM
  3. Simplifying Square roots
    Posted in the Algebra Forum
    Replies: 1
    Last Post: May 12th 2009, 01:22 PM
  4. Simplifying Square Roots
    Posted in the Algebra Forum
    Replies: 1
    Last Post: February 21st 2009, 12:11 PM
  5. Help with simplifying square roots
    Posted in the Algebra Forum
    Replies: 3
    Last Post: September 3rd 2007, 08:56 PM

Search Tags


/mathhelpforum @mathhelpforum