Elementary and Intermediate Algebra: Concepts & Applications (6th Edition)

Published by Pearson
ISBN 10: 0-32184-874-8
ISBN 13: 978-0-32184-874-1

Chapter 13 - Conic Sections - 13.1 Conic Sections: Parabolas and Circles - 13.1 Exercise Set - Page 855: 68



Work Step by Step

Consider expression $\sqrt{y}\sqrt[3]{{{y}^{2}}}$, Convert to the exponential notation, $\sqrt{y}\sqrt[3]{{{y}^{2}}}={{y}^{\frac{1}{2}}}\times {{y}^{\frac{2}{3}}}$ Add exponent, $\begin{align} & {{y}^{\frac{1}{2}+\frac{2}{3}}}={{y}^{\frac{3+4}{6}}} \\ & ={{y}^{^{\frac{7}{6}}}} \end{align}$ Convert back to radical notation, ${{y}^{^{\frac{7}{6}}}}=\sqrt[6]{{{y}^{7}}}$ Simplify the expression, $\sqrt[6]{{{y}^{7}}}=\sqrt[6]{{{y}^{6}}}\times \sqrt[6]{y}$ Find the sixth root. we assume $x\ge 0$, $y\sqrt[6]{y}$ Partial check: $\begin{matrix} {{\left( y\sqrt[6]{y} \right)}^{6}}\overset{?}{\mathop{=}}\,{{\left( y \right)}^{6}}{{\left( \sqrt[6]{y} \right)}^{6}} \\ \overset{?}{\mathop{=}}\,{{y}^{6}}\times y \\ ={{y}^{7}} \\ \end{matrix}$ Thus, the expression $\sqrt{y}\sqrt[3]{{{y}^{2}}}$ can be simplified as $y\sqrt[6]{y}$
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