# simplify the fractional expression

• Feb 6th 2008, 10:35 PM
cm3pyro
simplify the fractional expression
http://www.cmatros.com/9.jpg

I can't seem to get the correct answer for this question..
• Feb 7th 2008, 01:29 AM
earboth
Quote:

Originally Posted by cm3pyro
http://www.cmatros.com/9.jpg

I can't seem to get the correct answer for this question..

I'll show you step by step what I've done:

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

$\boxed{\frac{4x^6+1}{4x^3}}$
• Feb 7th 2008, 05:53 AM
cm3pyro
That's the answer that i got too, but according to my teacher, it's wrong.
• Feb 7th 2008, 06:18 AM
wingless
Quote:

Originally Posted by earboth
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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