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

  • #1
RChristenk
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5
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.
 
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  • #2
The original expression is positive, so there is no way to end up with negative values. If so, then it should be ##|y^3|##.
 
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  • #3
RChristenk said:
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?
 
  • #4
Mark44 said:
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##?
 
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  • #5
RChristenk said:
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.
RChristenk said:
Is there a way to simplify ##2x|x||y^2|y##?
That denominator should be ##2 x^2 |y^3|##, which is always positive.
 
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  • #6
FactChecker said:
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.
 
  • #7
RChristenk said:
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.
 
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