i have 2 more question that i tried to do but not sure do i have to go further. Thanks.

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- Sep 10th 2008, 01:44 PMxJenniferxRationalizing the Denominator #2
i have 2 more question that i tried to do but not sure do i have to go further. Thanks.

Link to problem.

ImageShack - Hosting :: 36817195gr4.png - Sep 10th 2008, 01:49 PMMoo
For the first one, you forgot $\displaystyle \sqrt{a-b}$ in the numerator again !

$\displaystyle \frac{\sqrt{a-b}\sqrt{a+b}}{a-b}=\frac{\sqrt{(a-b)(a+b)}}{a-b}=\frac{\sqrt{a^2-b^2}}{a-b}$

For the second one, assume that x,y>0 :

$\displaystyle \frac{\sqrt{x^3 y^5}}{x^3y^5}=\frac{\sqrt{x^2 y^4 xy}}{x^3y^5}=\frac{xy^2 \sqrt{xy}}{x^3y^5}=\dots$ - Sep 10th 2008, 01:51 PMxJenniferx
**Rationalizing the Denominator means taking the square root away....** - Sep 10th 2008, 01:57 PMShyam
$\displaystyle \frac{{\sqrt {a + b} }}

{{\sqrt {a - b} }} \hfill \\$

$\displaystyle = \frac{{\sqrt {a + b} }}

{{\sqrt {a - b} }} \times \frac{{\sqrt {a - b} }}

{{\sqrt {a - b} }} \hfill \\$

$\displaystyle = \frac{{\sqrt {\left( {a + b} \right)\left( {a - b} \right)} }}

{{a - b}} \hfill \\$

$\displaystyle = \frac{{\sqrt {a^2 - b^2 } }}

{{a - b}} \hfill \\

\hfill \\$

$\displaystyle {\text{Now, for second question}}{\text{.}} \hfill \\$

$\displaystyle \frac{1}

{{\sqrt {x^3 y^5 } }} = \frac{1}

{{\sqrt {x^2 xy^4 y} }} = \frac{1}

{{xy^2 \sqrt {xy} }} \hfill \\$

$\displaystyle = \frac{1}

{{xy^2 \sqrt {xy} }} \times \frac{{\sqrt {xy} }}

{{\sqrt {xy} }} \hfill \\$

$\displaystyle = \frac{{\sqrt {xy} }}

{{xy^2 xy}} = \frac{{\sqrt {xy} }}

{{x^2 y^3 }} \hfill \\

$ - Sep 10th 2008, 01:57 PMMoo