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: 43


The standard form of the expression $3\sqrt{-5}\left( -4\sqrt{-12} \right)$ is $24\sqrt{15}$.

Work Step by Step

Consider the expression,$3\sqrt{-5}\left( -4\sqrt{-12} \right)$ Express the square roots of negative numbers in terms of $i$. $\begin{align} & 3\sqrt{-5}\left( -4\sqrt{-12} \right)=3i\sqrt{5}\left( -4i\sqrt{12} \right) \\ & =-12{{i}^{2}}\sqrt{5}\left( \sqrt{12} \right) \end{align}$ Replace the value ${{i}^{2}}=-1$ and make the factors of the radicals. $\begin{align} & 3\sqrt{-5}\left( -4\sqrt{-12} \right)=-12\left( -1 \right)\sqrt{5}\left( \sqrt{4\cdot 3} \right) \\ & =12\sqrt{5}\left( 2\sqrt{3} \right) \\ & =24\sqrt{5}\cdot \sqrt{3} \end{align}$ Use the property $\sqrt{a}\cdot \sqrt{b}=\sqrt{ab}$. $\begin{align} & 3\sqrt{-5}\left( -4\sqrt{-12} \right)=24\sqrt{5\cdot 3} \\ & =24\sqrt{15} \end{align}$ Therefore, the standard form of the expression $3\sqrt{-5}\left( -4\sqrt{-12} \right)$ is $24\sqrt{15}$.
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