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\sqrt[4]{5x/8y}i asked my teacher and she said this "Hi there. separate the radicals - numerator and denominator first. Then, you'll need to make "8" into 2^3. You'll want to create a group of 4 for both the 2 and the y in the denominator, because your index is "4". So, you'll need one more "2" and "y^3". Then you multiply top and bottom by the 4th root of 2y^3 and simplify from there." can you show me what to do

User Floam
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Answer:
\frac{\sqrt[4]{10xy^3}}{2y}

where y is positive.

The 2y in the denominator is not inside the fourth root

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Work Shown:


\sqrt[4]{(5x)/(8y)}\\\\\\\sqrt[4]{(5x*2y^3)/(8y*2y^3)}\ \ \text{.... multiply top and bottom by } 2y^3\\\\\\\sqrt[4]{(10xy^3)/(16y^4)}\\\\\\\frac{\sqrt[4]{10xy^3}}{\sqrt[4]{16y^4}} \ \ \text{ ... break up the fourth root}\\\\\\\frac{\sqrt[4]{10xy^3}}{\sqrt[4]{(2y)^4}} \ \ \text{ ... rewrite } 16y^4 \text{ as } (2y)^4\\\\\\\frac{\sqrt[4]{10xy^3}}{2y} \ \ \text{... where y is positive}\\\\\\

The idea is to get something of the form
a^4 in the denominator. In this case,
a = 2y

To be able to reach the
16y^4, your teacher gave the hint to multiply top and bottom by
2y^3

For more examples, search out "rationalizing the denominator".

Keep in mind that
\sqrt[4]{(2y)^4} = 2y only works if y isn't negative.

If y could be negative, then we'd have to say
\sqrt[4]{(2y)^4} = |2y|. The absolute value bars ensure the result is never negative.

Furthermore, to avoid dividing by zero, we can't have y = 0. So all of this works as long as y > 0.

User Toongeorges
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