# Rationalizing the Denominator #2

• Sep 10th 2008, 01:44 PM
xJenniferx
Rationalizing the Denominator #2
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, 01:49 PM
Moo
Quote:

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

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For the first one, you forgot $\sqrt{a-b}$ in the numerator again !

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

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

$\frac{\sqrt{x^3 y^5}}{x^3y^5}=\frac{\sqrt{x^2 y^4 xy}}{x^3y^5}=\frac{xy^2 \sqrt{xy}}{x^3y^5}=\dots$
• Sep 10th 2008, 01:51 PM
xJenniferx
Rationalizing the Denominator means taking the square root away....
• Sep 10th 2008, 01:57 PM
Shyam
Quote:

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

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

$= \frac{{\sqrt {a + b} }}
{{\sqrt {a - b} }} \times \frac{{\sqrt {a - b} }}
{{\sqrt {a - b} }} \hfill \\$

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

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

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

$\frac{1}
{{\sqrt {x^3 y^5 } }} = \frac{1}
{{\sqrt {x^2 xy^4 y} }} = \frac{1}
{{xy^2 \sqrt {xy} }} \hfill \\$

$= \frac{1}
{{xy^2 \sqrt {xy} }} \times \frac{{\sqrt {xy} }}
{{\sqrt {xy} }} \hfill \\$

$= \frac{{\sqrt {xy} }}
{{xy^2 xy}} = \frac{{\sqrt {xy} }}
{{x^2 y^3 }} \hfill \\
$
• Sep 10th 2008, 01:57 PM
Moo
Quote:

Originally Posted by xJenniferx
Rationalizing the Denominator means taking the square root away....

Yes, but if you want to do so, you're likely to have square roots in the numerator. You can't just "forget" them ! :)