Results 1 to 6 of 6

Math Help - Simplifying radicals

  1. #1
    Member Jonboy's Avatar
    Joined
    May 2007
    From
    South Point
    Posts
    193

    Simplifying radicals

    Hello everyone! I would like it very much if you all checked my work. Thank you.

    My directions are: Simplify. Assume that no radicands were formed by raising a negative quantities to even powers.

    My problem is: \frac{\sqrt{128a^2b^4}}{\sqrt{16ab}}

    I dwindled it down to: \frac{\sqrt{64\cdot 2}ab^2}{4\sqrt{ab}}\;=\;\frac{8\sqrt{2}ab^2}{4\sqr  t{ab}}\;=\;\frac{2\sqrt{2}ab^2}{\sqrt{ab}}

    So now I need to rationalize the denominator right?

    So I did that and got: \boxed{\frac{2ab^2\sqrt{2ab}}{ab}}

    Does this look good, my math community?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    is up to his old tricks again! Jhevon's Avatar
    Joined
    Feb 2007
    From
    New York, USA
    Posts
    11,663
    Thanks
    3
    Quote Originally Posted by Jonboy View Post
    Hello everyone! I would like it very much if you all checked my work. Thank you.

    My directions are: Simplify. Assume that no radicands were formed by raising a negative quantities to even powers.

    My problem is: \frac{\sqrt{128a^2b^4}}{\sqrt{16ab}}

    I dwindled it down to: \frac{\sqrt{64\cdot 2}ab^2}{4\sqrt{ab}}\;=\;\frac{8\sqrt{2}ab^2}{4\sqr  t{ab}}\;=\;\frac{2\sqrt{2}ab^2}{\sqrt{ab}}

    So now I need to rationalize the denominator right?

    So I did that and got: \boxed{\frac{2ab^2\sqrt{2ab}}{ab}}

    Does this look good, my math community?
    i didn't check your working in detail, but it is obvious that your final answer is not simplified. you have ab in the denominator when we can cancel them with the ab^2 in the numerator.

    try this approach:

    \frac {\sqrt {128a^2b^4}}{\sqrt {16ab}} = \frac {128^{1/2}\left( a^2 \right)^{1/2}\left( b^4 \right)^{1/2}}{16^{1/2}a^{1/2}b^{1/2}} = \frac {128^{1/2}}{16^{1/2}} \cdot \frac {\left( a^2 \right)^{1/2}}{a^{1/2}} \cdot \frac {\left( b^4 \right)^{1/2}}{b^{1/2}}

    recall that \left( x^a \right)^b = x^{ab} and \frac {x^a}{x^b} = x^{a - b}

    simplify piece by piece, then recombine when done
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Member Jonboy's Avatar
    Joined
    May 2007
    From
    South Point
    Posts
    193
    Thanks Jhevon! I didn't see the ab cancle. So my final answer is 2b\sqrt{2ab}
    Follow Math Help Forum on Facebook and Google+

  4. #4
    is up to his old tricks again! Jhevon's Avatar
    Joined
    Feb 2007
    From
    New York, USA
    Posts
    11,663
    Thanks
    3
    Quote Originally Posted by Jonboy View Post
    Thanks Jhevon! I didn't see the ab cancle. So my final answer is 2b\sqrt{2ab}
    i don't like the fact that you have b's in different locations, i would write it as:

    2 \sqrt {2ab^3} or something like that
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Member Jonboy's Avatar
    Joined
    May 2007
    From
    South Point
    Posts
    193
    Is your way the most common? I just want that way.
    Follow Math Help Forum on Facebook and Google+

  6. #6
    is up to his old tricks again! Jhevon's Avatar
    Joined
    Feb 2007
    From
    New York, USA
    Posts
    11,663
    Thanks
    3
    Quote Originally Posted by Jonboy View Post
    Is your way the most common? I just want that way.
    i don't think it's that big a deal, it's just a preference. i think you've demonstrated the skills that the exercise was testing for
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Simplifying Radicals
    Posted in the Algebra Forum
    Replies: 5
    Last Post: October 23rd 2011, 02:28 PM
  2. Simplifying Radicals
    Posted in the Pre-Calculus Forum
    Replies: 1
    Last Post: November 25th 2009, 04:07 PM
  3. Simplifying Radicals
    Posted in the Algebra Forum
    Replies: 2
    Last Post: April 18th 2008, 07:27 PM
  4. #3 Simplifying Radicals
    Posted in the Algebra Forum
    Replies: 2
    Last Post: September 14th 2007, 04:49 PM
  5. simplifying radicals
    Posted in the Algebra Forum
    Replies: 3
    Last Post: July 24th 2005, 10:58 PM

Search Tags


/mathhelpforum @mathhelpforum