Algebra 2 Common Core

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

Chapter 4 - Quadratic Functions and Equations - 4-8 Complex Numbers - Practice and Problem-Solving Exercises - Page 254: 50



Work Step by Step

Since $\sqrt{ab}=\sqrt{a}\cdot\sqrt{b},$ the given expression, $ (4+\sqrt{-9})+(6-\sqrt{-49}) ,$ is equivalent to \begin{align*} & (4+\sqrt{-1}\cdot\sqrt{9})+(6-\sqrt{-1}\cdot\sqrt{49}) \\&= (4+\sqrt{-1}\cdot3)+(6-\sqrt{-1}\cdot7) \\&= (4+3\sqrt{-1})+(6-7\sqrt{-1}) .\end{align*} Since $i=\sqrt{-1},$ the expression above is equivalent to \begin{align*} (4+3i)+(6-7i) .\end{align*} Removing the grouping symbols, the expression above is equivalent to \begin{align*} 4+3i+6-7i .\end{align*} Combining the real parts and the imaginary parts, the expression above is equivalent to \begin{align*} & (4+6)+(3i-7i) \\&= 10+(-4i) \\&= 10-4i .\end{align*} Hence, the simplified form of the given expression is $ 10-4i $.
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