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 10 - Exponents and Radicals - Review Exercises: Chapter 10 - Page 694: 58



Work Step by Step

Multiplying by the conjugate of the denominator results to \begin{array}{l}\require{cancel} \dfrac{7-2i}{3+4i} \\\\= \dfrac{7-2i}{3+4i}\cdot\dfrac{3-4i}{3-4i} \\\\= \dfrac{(7-2i)(3-4i)}{(3+4i)(3-4i)} .\end{array} Using the product of the sum and difference of like terms which is given by $(a+b)(a-b)=a^2-b^2,$ the expression above is equivalent \begin{array}{l}\require{cancel} \dfrac{(7-2i)(3-4i)}{(3+4i)(3-4i)} \\\\= \dfrac{(7-2i)(3-4i)}{(3)^2-(4i)^2} \\\\= \dfrac{(7-2i)(3-4i)}{9-16i^2} .\end{array} Using the FOIL Method which is given by $(a+b)(c+d)=ac+ad+bc+bd,$ the given expression is equivalent to \begin{array}{l}\require{cancel} \dfrac{(7-2i)(3-4i)}{9-16i^2} \\\\= \dfrac{7(3)+7(-4i)-2i(3)-2i(-4i)}{9-16i^2} \\\\= \dfrac{21-28i-6i+8i^2}{9-16i^2} \\\\= \dfrac{21-34i+8i^2}{9-16i^2} .\end{array} Since $i^2=-1,$ the expression above is equivalent to \begin{array}{l}\require{cancel} \dfrac{21-34i+8i^2}{9-16i^2} \\\\= \dfrac{21-34i+8(-1)}{9-16(-1)} \\\\= \dfrac{21-34i-8}{9+16} \\\\= \dfrac{13-34i}{25} .\end{array}
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