Precalculus (6th Edition) Blitzer

Published by Pearson
ISBN 10: 0-13446-914-3
ISBN 13: 978-0-13446-914-0

Chapter 2 - Section 2.1 - Complex Numbers - Exercise Set - Page 314: 44


The standard form of the expression $3\sqrt{-7}\left( 2\sqrt{-8} \right)$ is $-12\sqrt{14}$.

Work Step by Step

Consider the expression,$3\sqrt{-7}\left( 2\sqrt{-8} \right)$ Express the square roots of negative numbers in terms of $i$. $\begin{align} & 3\sqrt{-7}\left( 2\sqrt{-8} \right)=3i\sqrt{7}\left( 2i\sqrt{8} \right) \\ & =6{{i}^{2}}\sqrt{7}\left( \sqrt{8} \right) \end{align}$ Replace the value ${{i}^{2}}=-1$ and make the factors of the radicals. $\begin{align} & 3\sqrt{-7}\left( 2\sqrt{-8} \right)=6\left( -1 \right)\sqrt{7}\left( \sqrt{4\cdot 2} \right) \\ & =-6\sqrt{7}\left( 2\sqrt{2} \right) \\ & =-12\sqrt{7}\cdot \sqrt{2} \end{align}$ Use the property $\sqrt{a}\cdot \sqrt{b}=\sqrt{ab}$. $\begin{align} & 3\sqrt{-7}\left( 2\sqrt{-8} \right)=-12\sqrt{7\cdot 2} \\ & =-12\sqrt{14} \end{align}$ Therefore, the standard form of the expression $3\sqrt{-7}\left( 2\sqrt{-8} \right)$ is $-12\sqrt{14}$.
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