# really easy cancel I can't do!

• May 5th 2008, 04:03 AM
i_zz_y_ill
really easy cancel I can't do!
Im not sure how to times through by 3/3 on the bottom

(2/3) / (1+(2/3)^2)^0.5

trying to get 2/(9+4x^2)^0.5 If somebody could show me how that'd be much appreciated
• May 5th 2008, 04:47 AM
colby2152
Quote:

Originally Posted by i_zz_y_ill
Im not sure how to times through by 3/3 on the bottom

(2/3) / (1+(2/3)^2)^0.5

trying to get 2/(9+4x^2)^0.5 If somebody could show me how that'd be much appreciated

What is $\displaystyle \frac{3}{3}$ squared? It is $\displaystyle \frac{9}{9}$.

Multiply that out through the "top" and "bottom" of the fraction.
• May 5th 2008, 04:48 AM
janvdl
Quote:

Originally Posted by i_zz_y_ill
Im not sure how to times through by 3/3 on the bottom

(2/3) / (1+(2/3)^2)^0.5

trying to get 2/(9+4x^2)^0.5 If somebody could show me how that'd be much appreciated

I'm sure you mean the following:

$\displaystyle \frac{ \frac{2}{3} }{ \sqrt{1 + \left( \frac{2x}{3} \right) ^{2}} }$

Observe the $\displaystyle \frac{2}{3}$ in the numerator. We can move that 3 to the denominator of the entire function.

Giving us:

$\displaystyle \frac{ 2 }{ 3 \sqrt{1 + \left( \frac{2x}{3} \right) ^{2}} }$

But we also know that:
$\displaystyle 3 = \sqrt{9}$

Thus:
$\displaystyle \frac{ 2 }{ \sqrt{9} \sqrt{1 + \left( \frac{2x}{3} \right) ^{2}} }$

Simplify the denominator.

$\displaystyle \frac{ 2 }{ \sqrt{9(1 + \left( \frac{2x}{3} \right) ^{2})} }$

$\displaystyle \frac{ 2 }{ \sqrt{9(1 + \left( \frac{4x^2}{9} \right) )} }$

$\displaystyle \frac{ 2 }{ \sqrt{9 + 4x^2 } }$