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Math Help - Confused - Radical Expression

  1. #1
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    Confused - Radical Expression

    Hello, everybody :]

    I'm a little confused as to the solution of a radical expression. My goal is to simplify it as much as I can. Here is the expression:

    <br />
\sqrt{\frac{9x^7}{16y^8}}<br />

    The solution which I thought would be right is:

    <br />
\frac{3x^6\sqrt{x}}{4y^7\sqrt{y}}<br />

    Using Mathway.com, however, it's telling me that the answer is:

    <br />
\frac{3x^2\sqrt{x}}{4y^4}<br />

    I would greatly appreciate if someone could explain how I got the wrong answer so I may learn from my mistakes. :] Thank you all very much.

    Colton
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  2. #2
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    Quote Originally Posted by Noegddgeon View Post
    Hello, everybody :]

    I'm a little confused as to the solution of a radical expression. My goal is to simplify it as much as I can. Here is the expression:

    <br />
\sqrt{\frac{9x^7}{16y^8}}<br />

    The solution which I thought would be right is:

    <br />
\frac{3x^6\sqrt{x}}{4y^7\sqrt{y}}<br />

    Using Mathway.com, however, it's telling me that the answer is:

    <br />
\frac{3x^2\sqrt{x}}{4y^4}<br />

    I would greatly appreciate if someone could explain how I got the wrong answer so I may learn from my mistakes. :] Thank you all very much.

    Colton
    recheck the "mathway.com" solution ...



    <br />
\sqrt{\frac{9x^7}{16y^8}} = \sqrt{\frac{3x^3 \cdot 3x^3 \cdot x}{4y^4 \cdot 4y^4}} = \sqrt{\frac{(3x^3)^2 \cdot x}{(4y^4)^2}}<br />
= \frac{\sqrt{(3x^3)^2} \cdot \sqrt{x}}{\sqrt{(4y^4)^2}} = \frac{3x^3 \sqrt{x}}{4y^4}
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  3. #3
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    Skeeter,

    Thank you for your reply. I must have entered the equation incorrectly into Mathway or something, because it gave me the one which I posted.... I find that rather strange. Thank you much, I appreciate your help.

    Colton
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  4. #4
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    I'm sorry to bump, but I have another question. Here I have the following problem I've stumbled on:

    <br />
\frac{x^{2/5}y^{5/6}}{x^{-1/3}y^{1/2}}<br />

    According to my workbook, the answer is x\sqrt[3]{y}

    However, I got the answer of xy\sqrt[3]{xy}

    My basis for solving the problem was to add the exponents on the top and bottom by finding the least common factor, which led me to simplify it to this:

    <br />
\frac{xy^{9/6}}{xy^{1/6}}<br />

    <br />
xy^{8/6}<br />

    <br />
xy^{4/3}<br />

    <br />
\sqrt[3]{xy^4}<br />

    <br />
xy\sqrt[3]{xy}<br />

    I am pretty much a beginner at math still, so I may have made a beginner's mistake when solving this problem. I would greatly appreciate some assistance, and I hope my step-by-step posting helps show where I may have made that mistake. :] Thank you very much in advance.

    Colton
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  5. #5
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    Quote Originally Posted by Noegddgeon View Post
    I'm sorry to bump, but I have another question. Here I have the following problem I've stumbled on:

    <br />
\frac{x^{2/\textcolor{red}{5}}y^{5/6}}{x^{-1/3}y^{1/2}}<br />

    According to my workbook, the answer is x\sqrt[3]{y}
    recheck the expression you typed ...

    you sure it's not \frac{x^{\frac{2}{\textcolor{red}{3}}} y^{\frac{5}{6}}}{x^{-\frac{1}{3}} y^{\frac{1}{2}}} ???

    if this is the original expression, then I agree w/ the workbook's solution.


    rule for division ...

    \frac{x^a}{x^b} = x^{a-b}


    x^{\frac{2}{3} - (-\frac{1}{3})} \cdot y^{\frac{5}{6} -  \frac{1}{2}}

    x^{1} \cdot y^{\frac{1}{3}}

    x \sqrt[3]{y}



    in future, please start a new problem with a new thread.
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  6. #6
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    Skeeter,

    Yes, that was a typo. My apologies. I greatly appreciate you showing me that problem, however, and I see my mistake now :] In the future, I will remember to start a new thread.

    Colton
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