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Math Help - Algebra glitch

  1. #1
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    Algebra glitch

    Hi,

    I'm not seeing the transition from 1 to 2. Could someone elucidate the matter for me, please?

    \bigg( \sqrt {2x^5 + 15x^\frac{4}{3} - 23x + 9} \bigg) \bigg( 4x^3  \bigg) + \bigg( x^4 \bigg) \frac{10x^4 + 20x^\frac{1}{3} - 23}{2 \sqrt  {2x^5 + 15x^\frac{4}{3} - 23x + 9}} ...\bigg( 1 \bigg)

    \equiv \frac{26x^8 + 140x\frac{13}{3} - 207x^4 + 72x^3}{2 \sqrt {2x^5 + 15x^\frac{4}{3} - 23x + 9}} ...\bigg( 2 \bigg)
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  2. #2
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    Start by creating a common denominator.
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  3. #3
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    I already tried that and ran into complications.

    \frac{2x^5 + 15x^\frac{4}{3} - 23x + 9)( Realised\, the\, futility\, of\, my\, attempt\, here)}{2 \sqrt {2x^5 + 15x^\frac{4}{3} - 23x + 9}}
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  4. #4
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    Quote Originally Posted by Hellbent View Post
    Hi,

    I'm not seeing the transition from 1 to 2. Could someone elucidate the matter for me, please?

    \bigg( \sqrt {2x^5 + 15x^\frac{4}{3} - 23x + 9} \bigg) \bigg( 4x^3  \bigg) + \bigg( x^4 \bigg) \frac{10x^4 + 20x^\frac{1}{3} - 23}{2 \sqrt  {2x^5 + 15x^\frac{4}{3} - 23x + 9}} ...\bigg( 1 \bigg)

    \equiv \frac{26x^8 + 140x\frac{13}{3} - 207x^4 + 72x^3}{2 \sqrt {2x^5 + 15x^\frac{4}{3} - 23x + 9}} ...\bigg( 2 \bigg)
    \displaystyle{4x^3\sqrt{2x^5 + 15x^{\frac{4}{3}} -23x + 9} + \frac{x^4(10x^4 + 20x^{\frac{1}{3}} - 23)}{2\sqrt{2x^5 + 15x^{\frac{4}{3}} - 23x + 9}} = \frac{8x^3(2x^5+ 15x^{\frac{4}{3}} - 23x + 9) + x^4(10x^4 + 20x^{\frac{1}{3}} - 23)}{2\sqrt{2x^5 + 15x^{\frac{4}{3}} - 23x+ 9}}}

    Now expand everything and simplify.
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  5. #5
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    Quote Originally Posted by Prove It View Post
    \displaystyle{4x^3\sqrt{2x^5 + 15x^{\frac{4}{3}} -23x + 9} + \frac{x^4(10x^4 + 20x^{\frac{1}{3}} - 23)}{2\sqrt{2x^5 + 15x^{\frac{4}{3}} - 23x + 9}} = \frac{8x^3(2x^5+ 15x^{\frac{4}{3}} - 23x + 9) + x^4(10x^4 + 20x^{\frac{1}{3}} - 23)}{2\sqrt{2x^5 + 15x^{\frac{4}{3}} - 23x+ 9}}}

    Now expand everything and simplify.
     \frac{16x^8 + 120x^\frac{13}{3} - 184x^4 + 72x^3 +10x^8 + 20x^\frac{13}{3} - 23x^4}{2\sqrt{2x^5 + 15x^{\frac{4}{3}} - 23x+ 9}}

    =  \frac{26x^8 + 140x\frac{13}{3} - 207x^4 + 72x^3}{2\sqrt{2x^5 + 15x^{\frac{4}{3}} - 23x+ 9}}

    Thank you!
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