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Math Help - Having difficulty understanding radicals

  1. #1
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    Having difficulty understanding radicals

    Will this cause me massive problems in the future if I don't fully 100% understand radicals? For example:

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

    I just don't know where to begin... I'm looking in the book for the answer and it says this.

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

    Then the next step: \sqrt[4]{\frac{24abc^2}{16b^4c^4}} Which I understand obviously, but then the final step is \frac{1}{2bc}\sqrt[4]{24abc^2}

    This is super complicated to me and it only gets worse... do I really need to know this?
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  2. #2
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    Re: Having difficulty understanding radicals

    Quote Originally Posted by uperkurk View Post
    Simplify \sqrt[4]{\frac{3a}{2b^3c^2}}
    \sqrt[4]{\frac{3a}{2b^3c^2}}=\sqrt[4]{\frac{3a}{2b^3c^2}\cdot\frac{8bc^2}{8bc^2}} why \color{red}8bc^2?
    Because you need a perfect 4^{th} power in that denominator: 2^4b^4c^4.
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    Re: Having difficulty understanding radicals

    Quote Originally Posted by uperkurk View Post
    Will this cause me massive problems in the future if I don't fully 100% understand radicals? For example:

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

    I just don't know where to begin... I'm looking in the book for the answer and it says this.

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

    Then the next step: \sqrt[4]{\frac{24abc^2}{16b^4c^4}} Which I understand obviously, but then the final step is \frac{1}{2bc}\sqrt[4]{24abc^2}

    This is super complicated to me and it only gets worse... do I really need to know this?
    2^4= 16. Since, in the original problem, there was a "2" in the denominator, they multiplied by 8 to get a "perfect fourth power". Similarly, b^3(b)= b^4 and c^2(c^2)= c^4. Multiplying the denominator by 8bc^2 makes the denominator 16b^4c^4= (2bc)^4 so that \sqrt[4]{(2bc)^4}= 2bc.

    Of course, if you multiply by 8bc^2 in the denominator you must do the same in the numerator.
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    Re: Having difficulty understanding radicals

    Quote Originally Posted by HallsofIvy View Post
    2^4= 16. Since, in the original problem, there was a "2" in the denominator, they multiplied by 8 to get a "perfect fourth power". Similarly, b^3(b)= b^4 and c^2(c^2)= c^4. Multiplying the denominator by 8bc^2 makes the denominator 16b^4c^4= (2bc)^4 so that \sqrt[4]{(2bc)^4}= 2bc.

    Of course, if you multiply by 8bc^2 in the denominator you must do the same in the numerator.
    Ok I see, but what if it was 3b^2c^4 for example?
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    Re: Having difficulty understanding radicals

    Quote Originally Posted by uperkurk View Post
    Ok I see, but what if it was 3b^2c^4 for example?
    Then use 27b^2 to get 3^4b^4c^4.
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  6. #6
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    Re: Having difficulty understanding radicals

    Oh right so you basically 3b^2c^4\cdot 27b^2 = 81b^4c^4 but then notice that 81 is 3^4 ok I see. It's pretty difficult because you need to think of a number that when multiplied, the 4th root will the original number, in our case 3.
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