Algebra 2 Common Core

Published by Prentice Hall
ISBN 10: 0133186024
ISBN 13: 978-0-13318-602-4

Chapter 6 - Radical Functions and Rational Exponents - 6-3 Binomial Radical Expressions - Practice and Problem-Solving Exercises - Page 379: 56



Work Step by Step

Extracting the root of the factor that is a perfect power of the index, the given expression, $ \dfrac{4+\sqrt{27}}{2-3\sqrt{27}} ,$ is equivalent to \begin{align*} & \dfrac{4+\sqrt{9\cdot3}}{2-3\sqrt{9\cdot3}} \\\\&= \dfrac{4+\sqrt{3^2\cdot3}}{2-3\sqrt{3^2\cdot3}} \\\\&= \dfrac{4+3\sqrt{3}}{2-3(3)\sqrt{3}} \\\\&= \dfrac{4+3\sqrt{3}}{2-9\sqrt{3}} .\end{align*} Multiplying the numerator and the denominator by the conjugate of the denominator, the expression above is equivalent to \begin{align*}\require{cancel} & \dfrac{4+3\sqrt{3}}{2-9\sqrt{3}}\cdot\dfrac{2+9\sqrt{3}}{2+9\sqrt{3}} \\\\&= \dfrac{(4+3\sqrt{3})(2+9\sqrt{3})}{(2)^2-(9\sqrt{3})^2} &\left( \text{use }(a+b)(a-b)=a^2-b^2 \right) \\\\&= \dfrac{(4+3\sqrt{3})(2+9\sqrt{3})}{4-81(3)} \\\\&= \dfrac{(4+3\sqrt{3})(2+9\sqrt{3})}{4-243} \\\\&= \dfrac{(4+3\sqrt{3})(2+9\sqrt{3})}{-239} \\\\&= \dfrac{4(2)+4(9\sqrt{3})+3\sqrt{3}(2)+3\sqrt{3}(9\sqrt{3})}{-239} &\left( \text{use FOIL} \right) \\\\&= \dfrac{8+36\sqrt{3}+6\sqrt{3}+27(3)}{-239} \\\\&= \dfrac{8+36\sqrt{3}+6\sqrt{3}+81}{-239} \\\\&= \dfrac{89+42\sqrt{3}}{-239} \\\\&= -\dfrac{89+42\sqrt{3}}{239} .\end{align*} Hence, the rationalized-denominator form of the given expression is $ -\dfrac{89+42\sqrt{3}}{239} $.
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