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 - Test: Chapter 10: 29

Answer

$\dfrac{-11-7i}{34}$

Work Step by Step

Multiplying by the conjugate of the denominator, the given expression is equivalent to \begin{array}{l}\require{cancel} \dfrac{-2+i}{3-5i} \\\\= \dfrac{-2+i}{3-5i}\cdot\dfrac{3+5i}{3+5i} \\\\= \dfrac{(-2+i)(3+5i)}{(3-5i)(3+5i)} .\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{(-2+i)(3+5i)}{(3-5i)(3+5i)} \\\\= \dfrac{(-2+i)(3+5i)}{(3)^2-(5i)^2} \\\\= \dfrac{(-2+i)(3+5i)}{9-25i^2} .\end{array} Using the FOIL Method which is given by $(a+b)(c+d)=ac+ad+bc+bd,$ the expression above is equivalent to\begin{array}{l}\require{cancel} \dfrac{-2(3)-2(5i)+i(3)+i(5i)}{9-25i^2} \\\\= \dfrac{-6-10i+3i+5i^2}{9-25i^2} \\\\= \dfrac{-6-7i+5i^2}{9-25i^2} .\end{array} Since $i^2=-1,$ the expression above is equivalent to \begin{array}{l}\require{cancel} \dfrac{-6-7i+5i^2}{9-25i^2} \\\\= \dfrac{-6-7i+5(-1)}{9-25(-1)} \\\\= \dfrac{-6-7i-5}{9+25} \\\\= \dfrac{-11-7i}{34} .\end{array}
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