Rationalize the denominator: ##\sqrt{\dfrac{1}{2x^3y^5}}##

[RChristenk]

Homework Statement

Rationalize the denominator: $\sqrt{\dfrac{1}{2x^3y^5}}$

Relevant Equations

Operations involving radicals

$\sqrt{\dfrac{1}{2x^3y^5}}=\dfrac{1}{\sqrt{2\cdot x^2 \cdot x \cdot y^2 \cdot y^2 \cdot y}}=\dfrac{1}{|x|\cdot |y|\cdot |y| \cdot \sqrt{2xy}}=\dfrac{1}{|x|y^2\sqrt{2xy}}$

$\Rightarrow \dfrac{1}{|x|y^2\sqrt{2xy}} \cdot \dfrac{\sqrt{2xy}}{\sqrt{2xy}}=\dfrac{\sqrt{2xy}}{|x|y^2 \cdot 2xy}=\dfrac{\sqrt{2xy}}{2|x|xy^3}$

But the solution is given as $\sqrt{\dfrac{1}{2x^3y^5}}=\dfrac{1}{xy^2\sqrt{2xy}}\cdot \dfrac{\sqrt{2xy}}{\sqrt{2xy}}=\dfrac{\sqrt{2xy}}{2x^2y^3}$ without any consideration for the absolute value. But the definition is $\sqrt{x^2}=|x|$, so I'm not understanding why the book solution ignores this.

 

[fresh_42]

The original expression is positive, so there is no way to end up with negative values. If so, then it should be $|y^3|$.

 

[Mark44]

I'm not understanding why the book solution ignores this.

Are there assumptions in the problem that you don't show? IOW, are the variables x and y assumed to be nonnegative?

 

[RChristenk]

Are there assumptions in the problem that you don't show? IOW, are the variables x and y assumed to be nonnegative?

Nope. But if $x,y$ can be negative, would this be correct: $\dfrac{1}{|x||y^2|\sqrt{2xy}}\cdot \dfrac{\sqrt{2xy}}{\sqrt{2xy}}=\dfrac{\sqrt{2xy}}{2x|x||y^2|y}$.

Is there a way to simplify $2x|x||y^2|y$?

 

[FactChecker]

Nope. But if $x,y$ can be negative, would this be correct: $\dfrac{1}{|x||y^2|\sqrt{2xy}}\cdot \dfrac{\sqrt{2xy}}{\sqrt{2xy}}=\dfrac{\sqrt{2xy}}{2x|x||y^2|y}$.

No, because your final answer would be negative if exactly one of x or y were negative. The original fraction is always positive.

Is there a way to simplify $2x|x||y^2|y$?

That denominator should be $2 x^2 |y^3|$, which is always positive.

 

[RChristenk]

No, because your final answer would be negative if exactly one of x or y were negative. The original fraction is always positive.

That denominator should be $2 x^2 |y^3|$, which is always positive.

Why is the original fraction always positive? I'm confused because by definition $\sqrt{x^2}=|x|$. So $x$ itself could be a negative number.

Hence in $\sqrt{\dfrac{1}{2x^3y^5}}$, $x,y$ could be negative and since it's to the third and fifth power respectively, stay negative. Thanks.

 

[FactChecker]

Why is the original fraction always positive? I'm confused because by definition $\sqrt{x^2}=|x|$. So $x$ itself could be a negative number.

Hence in $\sqrt{\dfrac{1}{2x^3y^5}}$, $x,y$ could be negative and since it's to the third and fifth power respectively, stay negative. Thanks.

I'm sorry. I misstated. $\sqrt {\frac {1}{2x^3y^5}}$ should always be positive, when it is defined. (I am assuming that your class does not deal with complex roots of negative numbers.) That is how the $\sqrt {\ \ \ }$ symbol is defined.