1. ## Algebra glitch

Hi,

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

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

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

2. Start by creating a common denominator.

3. I already tried that and ran into complications.

$\frac{2x^5 + 15x^\frac{4}{3} - 23x + 9)( Realised\, the\, futility\, of\, my\, attempt\, here)}{2 \sqrt {2x^5 + 15x^\frac{4}{3} - 23x + 9}}$

4. Originally Posted by Hellbent
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.

5. Originally Posted by Prove It
$\displaystyle{4x^3\sqrt{2x^5 + 15x^{\frac{4}{3}} -23x + 9} + \frac{x^4(10x^4 + 20x^{\frac{1}{3}} - 23)}{2\sqrt{2x^5 + 15x^{\frac{4}{3}} - 23x + 9}} = \frac{8x^3(2x^5+ 15x^{\frac{4}{3}} - 23x + 9) + x^4(10x^4 + 20x^{\frac{1}{3}} - 23)}{2\sqrt{2x^5 + 15x^{\frac{4}{3}} - 23x+ 9}}}$

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!