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Math Help - simplify the fractional expression

  1. #1
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    simplify the fractional expression



    I can't seem to get the correct answer for this question..
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  2. #2
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    Quote Originally Posted by cm3pyro View Post


    I can't seem to get the correct answer for this question..
    I'll show you step by step what I've done:

    \sqrt{1+\left(x^3-\frac1{4x^3}\right)^2} = \sqrt{1+\frac{(4x^6-1)^2}{16x^6}} = \frac1{4x^3} \cdot \sqrt{16x^6+(4x^6-1)^2}  = \frac1{4x^3} \cdot \sqrt{16x^6+16x^{12} - 8x^6 + 1} = \frac1{4x^3} \cdot \sqrt{16x^{12} + 8x^6 + 1} = \frac1{4x^3} \cdot \sqrt{(4x^6+1)^2} =

    \boxed{\frac{4x^6+1}{4x^3}}
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  3. #3
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    That's the answer that i got too, but according to my teacher, it's wrong.
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  4. #4
    Super Member wingless's Avatar
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    Quote Originally Posted by earboth View Post
    I'll show you step by step what I've done:

    \sqrt{1+\left(x^3-\frac1{4x^3}\right)^2} = \sqrt{1+\frac{(4x^6-1)^2}{16x^6}} = \frac1{4x^3} \cdot \sqrt{16x^6+(4x^6-1)^2}  = \frac1{4x^3} \cdot \sqrt{16x^6+16x^{12} - 8x^6 + 1} = \frac1{4x^3} \cdot \sqrt{16x^{12} + 8x^6 + 1} = \frac1{4x^3} \cdot \sqrt{(4x^6+1)^2} =

    \boxed{\frac{4x^6+1}{4x^3}}
    The most common mistake in algebra calculations, \sqrt{x^2} \neq x, \sqrt{x^2} = |x|!

    So you're right until \frac{\sqrt{(4x^6+1)^2}}{\sqrt{(4x^3)^2}}

    It should be
    \frac{|4x^6+1|}{|4x^3|} =  \left | \frac{4x^6+1}{4x^3} \right |
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