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Math Help - Reducing Roots

  1. #1
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    Reducing Roots

    Expression \sqrt[3]{16x^7y^5}

    1. \sqrt[3]{2^32x^3x^3xy^3y^2}
    -I understand simplyifying in this step. \sqrt[3]{16} = 2^3*2

    2. \sqrt[3]{2^3} \sqrt[3]{x^3} \sqrt[3]{x^3} \sqrt[3]{y^3} \sqrt[3]{2xy^2}
    -Here is where im confused. How and why are you able to separate the 2^3*2 from \sqrt[3]{16} into two separate locations? At the begging of the expresion the 2^3 is present, while the other 2 is all the way at the end of the expression with \sqrt[3]{2xy^2}.

    Many thanks for clarifying!
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  2. #2
    MHF Contributor Amer's Avatar
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    Quote Originally Posted by allyourbass2212 View Post
    Expression \sqrt[3]{16x^7y^5}

    1. \sqrt[3]{2^32x^3x^3xy^3y^2}
    -I understand simplyifying in this step. \sqrt[3]{16} = 2^3*2

    2. \sqrt[3]{2^3} \sqrt[3]{x^3} \sqrt[3]{x^3} \sqrt[3]{y^3} \sqrt[3]{2xy^2}
    -Here is where im confused. How and why are you able to separate the 2^3*2 from \sqrt[3]{16} into two separate locations? At the begging of the expresion the 2^3 is present, while the other 2 is all the way at the end of the expression with \sqrt[3]{2xy^2}.

    Many thanks for clarifying!
    look
    you know that

    \sqrt[3]{ab}=(\sqrt[3]{a})(\sqrt[3]{b})

    2(4)=\sqrt{4}(4)=\sqrt{4}(\sqrt{16})=\sqrt{4(16)}=  \sqrt{64}=8

    \sqrt[3]{(2)(2)(2)}=2

    the square degree is 3 so I take one thing from every three thing

    }=\sqrt[3]{}" alt="\sqrt[3]{}=\sqrt[3]{}" />

    \sqrt[4]{{\color{red}2*2*2*2}*2*{\color{blue}3*3*3*3}}=2(3  )\sqrt[4]{2}

    when I said

    2^4=2(2)(2)(2) four times

    take the fourth sqrt

    \sqrt[4]{(2)(2)(2)(2)}=2

    if

    \sqrt{2(2)(2)}=2\sqrt{2}

    when we said the square of degree four of A that's mean
    we need a number if you multiply itself by itself four times we get A

    it is clear or not
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  3. #3
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    I understand your examples, but im sorry I do not grasp whats going on from here   \sqrt[3]{2^32x^3x^3xy^3y^2} to   \sqrt[3]{2^3} \sqrt[3]{x^3} \sqrt[3]{x^3} \sqrt[3]{y^3} \sqrt[3]{2xy^2}. Specifically regarding the movement of the 2 from the beginning to the end of the expression.

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  4. #4
    MHF Contributor Amer's Avatar
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    Quote Originally Posted by allyourbass2212 View Post
    I understand your examples, but im sorry I do not grasp whats going on from here   \sqrt[3]{2^32x^3x^3xy^3y^2} to   \sqrt[3]{2^3} \sqrt[3]{x^3} \sqrt[3]{x^3} \sqrt[3]{y^3} \sqrt[3]{2xy^2}. Specifically regarding the movement of the 2x to the end of the expression.

    you know that

    abcd=acbd=bacd=dacb

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

    lats simplify it
    you know that y^5=(y)(y)(y)(y)(y)
    x^7=x(x)(x)(x)(x)(x)(x)
    16=2(2)(2)(2)
    for each three same thing take one



    \sqrt[3]{16y^5x^7}=\sqrt[3]{2{\color{red}(2)(2)(2)}(y)(y){\color{blue}(7)(7)(  7)}x{\color{green}(x)(x)(x)}{\color{red}(x)(x)(x)}  }
    you will take two x since you have 6 x's and one y since you have three y and one 2 since you have three what will stay in the sqrt

    (x)(x)(y)(2)\sqrt[3]{2(y)(y)(x)}

    now collect them
    like this
    (x)(x)=x^2

    2x^2(y)\sqrt[3]{2y^2x}

    you can change the order since the multiple is a commutative operation

    2(y)x^2\sqrt[3]{2xy^2}

    I think it is clear
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  5. #5
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    Thanks Amer, that method makes sense!

    I am not sure what the book is trying to do but thats not important at the moment.

    The only question I have is how do you know \sqrt[3]{16}=
    (2)(2)(2)(2)

    And could you list the break down of some other numbers for example?

    Thanks!
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  6. #6
    MHF Contributor Amer's Avatar
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    Quote Originally Posted by allyourbass2212 View Post
    Thanks Amer, that method makes sense!

    I am not sure what the book is trying to do but thats not important at the moment.

    The only question I have is how do you know \sqrt[3]{16}=
    (2)(2)(2)(2)


    And could you list the break down of some other numbers for example?

    Thanks!
    ok sure

    \sqrt[3]{16}= \sqrt[3]{2{\color{red}(2)(2)(2)}}=2\sqrt[3]{2}

    easy one

    \sqrt{144}=\sqrt{2(72)}=\sqrt{{\color{red}2(2)}(36  )}=2\sqrt{36}=2\sqrt{2(2)(3)(3)}=2(2)(3)=12

    level is medium

    \sqrt[3]{864}=\sqrt[3]{2(432)}=\sqrt[3]{2(2)(216)};=\sqrt[3]{{\color{red}(2)(2)(2)}(108)}=2\sqrt[3]{108}=
    2\sqrt[3]{2(54)}=2\sqrt[3]{2(2)(27)}=2\sqrt[3]{(2)(2)(3)(9)}=2\sqrt[3]{(2)(2){\color{blue}(3)(3)(3)}}=2(3)\sqrt[3]{2^2}

    hard one here I will not expand like before I will write  a(a)(a)=a^3 right


    \sqrt[4]{y^15x^5z^7}=\sqrt[4]{{\color{blue}(y^4)(y^4)(y^4)}(y^3){\color{red}(x^  4)}(x){\color{red}(z^4)}(z^3)} =y(y)(y)(x)(z)\sqrt[4]{y^3(x)(z^3)}=y^3(x)(z)\sqrt[4]{y^3(x)(z^3)}

    if there is any question feel free to post it .........
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