College Algebra (10th Edition)

Published by Pearson
ISBN 10: 0321979478
ISBN 13: 978-0-32197-947-6

Chapter 1 - Section 1.3 - Complex Numbers; Quadratic Equations in the Complex Number System - 1.3 Assess Your Understanding: 3

Answer

$6-3\sqrt3$

Work Step by Step

Rationalize the denominator by multiplying the conjugate of the denominator (which is $2-\sqrt3$) to both the numerator and the denominator to obtain: $=\dfrac{3(2-\sqrt3)}{(2+\sqrt3)(2-\sqrt3)} \\=\dfrac{6-3\sqrt3}{(2+\sqrt3)(2-\sqrt3)}$ Use the rule $(a-b)(a+b)=a^2-b^2$ to obtain: $=\dfrac{6-3\sqrt3}{2^2-(\sqrt3)^2} \\=\dfrac{6-3\sqrt3}{4-3} \\=\dfrac{6-3\sqrt3}{1} \\=6-3\sqrt3$
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