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Math Help - Simplifying, its easy.

  1. #1
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    Simplifying, its easy.

    (100 / 4 root x) - (36 /x^2)

    I get this far:

    (10^2 / x^(1/2)) - (6^2 /x^2)

    EDIT: Anyone have any idea?
    Last edited by Jeavus; September 26th 2007 at 01:43 PM.
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  2. #2
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    Quote Originally Posted by Jeavus View Post
    (100 / 4 root x) - (36 /x^2)

    I get this far:

    (10^2 / x^(1/2)) - (6^2 /x^2)
    (You missed a 4 in the first fraction, by the way.)

    You have two options as I see it.

    1. Add the fractions.
    Find a common denominator. The LCM of 4x^{1/2} \text{ and }x^2 is 4x^2.

    So
    \frac{100}{x^{1/2}} \cdot \frac{x^{3/2}}{x^{3/2}} - \frac{36}{x^2} \cdot \frac{4}{4}

    = \frac{100x^{3/2} - 144}{4x^2} = \frac{25x^{3/2} - 72}{x^2}

    2. Simplify each term independently.
    \frac{100}{4\sqrt{x}} - \frac{36}{x^2}

    = \frac{25}{\sqrt{x}} - \frac{36}{x^2}

    = \frac{25 \sqrt{x}}{x} - \frac{36}{x^2}

    Now factor:
    = \frac{1}{x} \cdot \left ( 25 - \frac{36}{x} \right )

    -Dan
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  3. #3
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    I think you misinterpreted the question.

    4 root x

    It looks like this:

    http://img255.imageshack.us/img255/8985/4rootuh6.png

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  4. #4
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by Jeavus View Post
    I think you misinterpreted the question.

    4 root x

    It looks like this:

    http://img255.imageshack.us/img255/8985/4rootuh6.png

    Ahhhhhh....

    Hmmmm.... It doesn't seem like we'd want to add the fractions here, so I would simplify the first term, then factor:
    \frac{100}{\sqrt[4]{x}} - \frac{36}{x^2}

    = \frac{100}{\sqrt[4]{x}} \cdot \frac{\sqrt[4]{x^3}}{\sqrt[4]{x^3}} - \frac{36}{x^2}

    = \frac{100 \sqrt[4]{x^3}}{x} - \frac{36}{x^2}

    = \frac{1}{x} \left ( 100 \sqrt[4]{x^3} - \frac{36}{x} \right )

    -Dan
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  5. #5
    Forum Admin topsquark's Avatar
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    I suppose you could also treat this as the difference between two squares:
    \frac{100}{x^{1/4}} - \frac{36}{x^2}

    = \left ( \frac{10}{x^{1/8}} + \frac{6}{x} \right )\left ( \frac{10}{x^{1/8}} - \frac{6}{x} \right )

    = 4 \left ( \frac{5}{x^{1/8}} + \frac{3}{x} \right )\left ( \frac{5}{x^{1/8}} - \frac{3}{x} \right )

    -Dan
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    How does it become 10 / x^1/8?
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  7. #7
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by Jeavus View Post
    How does it become 10 / x^1/8?
    Recall that a^2 - b^2 = (a + b)(a - b).

    So your question becomes: How do you take the square root of \frac{100}{x^{1/4}}?

    I assume you can do the square root of 100, so how do you do the square root of x^{1/4}?
    \sqrt{x^{1/4}} = \left ( x^{1/4} \right ) ^{1/2} = x^{1/4 \cdot 1/2} = x^{1/8}

    Thus
    \sqrt{\frac{100}{x^{1/4}}} = \frac{10}{x^{1/8}}

    -Dan
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