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Simplify $\frac{3}{\sqrt[5]{16}}+\frac{1}{\sqrt{3}}$ and rationalize the denominator. The result can be expressed in the form $\frac{a^2\sqrt[5]{b}+b\sqrt{a}}{ab}$, where $a$ and $b$ are integers. What is the value of the sum $a+b$?

 
Guest Jan 13, 2018

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 #1
avatar+5924 
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\(\quad\,\frac{3}{\sqrt[5]{16}}\,+\,\frac{1}{\sqrt3} \\~\\ =\,\frac{3}{\sqrt[5]{2^4}}\,+\,\frac{1}{\sqrt3} \\~\\ =\,\frac{3}{2^\frac45}\,+\,\frac1{3^\frac12}\\~\\ =\,\frac{3\,\cdot\,2^\frac15}{2^\frac45\,\cdot\,2^\frac15}\,+\,\frac{1\,\cdot\,3^\frac12}{3^\frac12\,\cdot\,3^\frac12}\\~\\ =\,\frac{3\,\cdot\,2^\frac15}{2}\,+\,\frac{3^\frac12}{3}\\~\\ =\,\frac{3\,\cdot\,2^\frac15\,\cdot\,3}{2\,\cdot\,3}\,+\,\frac{3^\frac12\,\cdot\,2}{3\,\cdot\,2}\\~\\ =\,\frac{3^2\,\cdot\,2^\frac15}{3\cdot2}\,+\,\frac{3^\frac12\,\cdot\,2}{3\cdot2}\\~\\ =\,\frac{3^2\,\cdot\,2^\frac15\,+\,3^\frac12\,\cdot\,2}{3\cdot2}\\~\\ =\,\frac{3^2\sqrt[5]{2}\,+\,2\sqrt3}{3\cdot2}\)

 

a = 3    and    b = 2     so     a + b  =  3 + 2  =  5

 
hectictar  Jan 13, 2018
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 #1
avatar+5924 
+1
Best Answer

\(\quad\,\frac{3}{\sqrt[5]{16}}\,+\,\frac{1}{\sqrt3} \\~\\ =\,\frac{3}{\sqrt[5]{2^4}}\,+\,\frac{1}{\sqrt3} \\~\\ =\,\frac{3}{2^\frac45}\,+\,\frac1{3^\frac12}\\~\\ =\,\frac{3\,\cdot\,2^\frac15}{2^\frac45\,\cdot\,2^\frac15}\,+\,\frac{1\,\cdot\,3^\frac12}{3^\frac12\,\cdot\,3^\frac12}\\~\\ =\,\frac{3\,\cdot\,2^\frac15}{2}\,+\,\frac{3^\frac12}{3}\\~\\ =\,\frac{3\,\cdot\,2^\frac15\,\cdot\,3}{2\,\cdot\,3}\,+\,\frac{3^\frac12\,\cdot\,2}{3\,\cdot\,2}\\~\\ =\,\frac{3^2\,\cdot\,2^\frac15}{3\cdot2}\,+\,\frac{3^\frac12\,\cdot\,2}{3\cdot2}\\~\\ =\,\frac{3^2\,\cdot\,2^\frac15\,+\,3^\frac12\,\cdot\,2}{3\cdot2}\\~\\ =\,\frac{3^2\sqrt[5]{2}\,+\,2\sqrt3}{3\cdot2}\)

 

a = 3    and    b = 2     so     a + b  =  3 + 2  =  5

 
hectictar  Jan 13, 2018

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