Results 1 to 5 of 5

Math Help - Roots Need Help

  1. #1
    Member cmf0106's Avatar
    Joined
    May 2008
    Posts
    104

    Roots Need Help

    This section talks about how you can simplify roots by combining two properties, ill refer to as A & B.

    in {^n}\sqrt{a^m} if m is at least as large as n then this can be simplified using

    Property A. ({^n}\sqrt{ab})={^n}\sqrt{a}  {^n\sqrt{b}}
    &
    Property B. {^n}\sqrt{a^n} = a

    then goes on to list this as an example.

    \sqrt{32x^3}=\sqrt{2^22^22^1x^2x^1}= \sqrt{2^2} \sqrt{2^2} \sqrt{x^2} \sqrt{2x} = 2 * 2x\sqrt{2x} = 4x\sqrt{2x}

    I have no idea whats going on here. If someone could please walk me through this it would be greatly appreciated. Many thanks.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    A riddle wrapped in an enigma
    masters's Avatar
    Joined
    Jan 2008
    From
    Big Stone Gap, Virginia
    Posts
    2,551
    Thanks
    12
    Awards
    1
    Quote Originally Posted by cmf0106 View Post
    This section talks about how you can simplify roots by combining two properties, ill refer to as A & B.

    in {^n}\sqrt{a^m} if m is at least as large as n then this can be simplified using

    Property A. ({^n}\sqrt{ab})={^n}\sqrt{a} {^n\sqrt{b}}
    &
    Property B. {^n}\sqrt{a^n} = a

    then goes on to list this as an example.

    \sqrt{32x^3}=\sqrt{2^22^22^1x^2x^1}= \sqrt{2^2} \sqrt{2^2} \sqrt{x^2} \sqrt{2x} = 2 * 2x\sqrt{2x} = 4x\sqrt{2x}

    I have no idea whats going on here. If someone could please walk me through this it would be greatly appreciated. Many thanks.
    Let me show another way to look at it. See if you like it.

    \sqrt{32x^3}

    First thing I do is change everything in the radicand to exponents:

    \sqrt{2^5x^3}

    Now divide the each exponent of each factor in the radicand by the index of the radical, in this case (2):

    \frac{5}{2}=2 \ \ remainder \ \ 1

    \frac{3}{2}=1 \ \ remainder \ \ 1

    This means that a 2^2 is extracted from the radicand and a 2^1 remains.  x^1 is extracted from the radicand and x^1 remains. What's extracted is 4x and what's left in the radicand is 2x.

    \sqrt{2^5x^3}=2^2x\sqrt{2x}=4x\sqrt{2x}
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Member cmf0106's Avatar
    Joined
    May 2008
    Posts
    104
    EDIT: Double checked, I think that makes sense. Let me try your method on an additional problem see if I have the right idea.

    {^3}\sqrt{8y}^3

    1. Change To Exponents \sqrt{2^3y^3}

    2. Divide each exponent of each factor by the index of the radical in this case 3

    3\div3=1no remains
    3\div3=1no remains

    3. 2y for your final answer.

    &

    {^3}\sqrt{16x^7y^5}

    1. Change to Exponents
    \sqrt{2^4x^7y^5}

    2.
    (2^4) = 4\div3 = 1 & 1 remains
    (x^7) = 2 and 1 remains
    (y^5) = 1 and 2 remains

    3. 2x^2y^3 {^3}\sqrt{2xy^2}

    Do these look right?
    And it seems that two criteria always apply
    1. M is at least as large as N
    2. in your final answer if there is a radicand it will always be to the originals index
    Last edited by cmf0106; August 17th 2008 at 01:57 PM.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    A riddle wrapped in an enigma
    masters's Avatar
    Joined
    Jan 2008
    From
    Big Stone Gap, Virginia
    Posts
    2,551
    Thanks
    12
    Awards
    1
    Quote Originally Posted by cmf0106 View Post
    EDIT: Double checked, That makes more sense up until where you state 2^2 is extracted from the radicand. Shouldn't it be 2^1 since 5/2= 2 remainder 1?
    No.

    \sqrt{2^5x^3}=2^{\frac{5}{2}}x^{\frac{3}{2}}=2^2 \cdot 2^{\frac{1}{2}} \cdot x^1 \cdot x^{\frac{1}{2}}=4x \cdot 2^{\frac{1}{2}} \cdot x^{\frac{1}{2}}=4x\sqrt{2x}
    Follow Math Help Forum on Facebook and Google+

  5. #5
    A riddle wrapped in an enigma
    masters's Avatar
    Joined
    Jan 2008
    From
    Big Stone Gap, Virginia
    Posts
    2,551
    Thanks
    12
    Awards
    1
    Quote Originally Posted by cmf0106 View Post
    EDIT: Double checked, I think that makes sense. Let me try your method on an additional problem see if I have the right idea.

    {^3}\sqrt{8y}^3

    1. Change To Exponents \sqrt{2^3y^3}

    2. Divide each exponent of each factor by the index of the radical in this case 3

    3\div3=1no remains
    3\div3=1no remains

    3. 2y for your final answer.
    Correctamundo!!
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. roots
    Posted in the Advanced Algebra Forum
    Replies: 2
    Last Post: December 24th 2010, 02:50 AM
  2. Roots & Imaginary Roots
    Posted in the Math Topics Forum
    Replies: 4
    Last Post: October 4th 2009, 09:24 AM
  3. roots
    Posted in the Advanced Algebra Forum
    Replies: 3
    Last Post: April 29th 2009, 05:27 AM
  4. Roots ><
    Posted in the Algebra Forum
    Replies: 3
    Last Post: October 7th 2008, 08:49 AM
  5. Given 2 imaginary roots find other 2 roots
    Posted in the Pre-Calculus Forum
    Replies: 1
    Last Post: January 26th 2008, 09:24 PM

Search Tags


/mathhelpforum @mathhelpforum