Question

ign="absmiddle" class="latex-formula">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

Answer 1

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:

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

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.