College Algebra (11th Edition)

Published by Pearson
ISBN 10: 0321671791
ISBN 13: 978-0-32167-179-0

Chapter R - Section R.7 - Radical Expressions - R.7 Exercises - Page 68: 88


$\dfrac{\sqrt[4]{7}+t^2\sqrt[]{3}}{t^{3}} $

Work Step by Step

$\bf{\text{Solution Outline:}}$ Simplify each term in the given expression, $ \sqrt[4]{\dfrac{7}{t^{12}}}+\sqrt[4]{\dfrac{9}{t^4}} .$ Then make the terms similar (same denominator) to combine the numerators. $\bf{\text{Solution Details:}}$ Simplifying each term of the expression above results to \begin{array}{l}\require{cancel} \sqrt[4]{\dfrac{1}{t^{12}}\cdot7}+\sqrt[4]{\dfrac{1}{t^4}\cdot9} \\\\= \sqrt[4]{\left(\dfrac{1}{t^{3}}\right)^4\cdot7}+\sqrt[4]{\left(\dfrac{1}{t}\right)^4\cdot9} \\\\= \dfrac{1}{t^{3}}\sqrt[4]{7}+\dfrac{1}{t}\sqrt[4]{9} \\\\= \dfrac{\sqrt[4]{7}}{t^{3}}+\dfrac{\sqrt[4]{9}}{t} .\end{array} Using the definition of rational exponents which is given by $a^{\frac{m}{n}}=\sqrt[n]{a^m}=\left(\sqrt[n]{a}\right)^m,$ the expression $\sqrt[4]{9}$ above is equivalent to \begin{array}{l}\require{cancel} 9^{\frac{1}{4}} \\\\= (3^2)^{\frac{1}{4}} \\\\= 3^{2\cdot\frac{1}{4}} \\\\= 3^{\frac{2}{4}} \\\\= 3^{\frac{1}{2}} \\\\= \sqrt{3} .\end{array} Hence, the expression, $\dfrac{\sqrt[4]{7}}{t^{3}}+\dfrac{\sqrt[4]{9}}{t},$ is equivalent to \begin{array}{l}\require{cancel} \dfrac{\sqrt[4]{7}}{t^{3}}+\dfrac{\sqrt[]{3}}{t} .\end{array} To simplify the expression above, make the terms similar by multiplying the necessary term/s to an expression equal to $1$. Hence, the expression above is equivalent to \begin{array}{l}\require{cancel} \dfrac{\sqrt[4]{7}}{t^{3}}+\dfrac{\sqrt[]{3}}{t}\cdot\dfrac{t^2}{t^2} \\\\= \dfrac{\sqrt[4]{7}}{t^{3}}+\dfrac{t^2\sqrt[]{3}}{t^3} .\end{array} To combine similar terms, add/subtract the numerators and copy the similar denominator. Hence, the expression above is equivalent to \begin{array}{l}\require{cancel} \dfrac{\sqrt[4]{7}+t^2\sqrt[]{3}}{t^{3}} .\end{array}
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