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 - Page 695: 19



Work Step by Step

Isolating the radical expression, the given expression is equivalent to \begin{array}{l}\require{cancel} 6=\sqrt{x-3}+5 \\\\ 6-5=\sqrt{x-3} \\\\ 1=\sqrt{x-3} .\end{array} Squaring both sides of the equation and then isolating the variable, the expression above is equivalent to \begin{array}{l}\require{cancel} 1=\sqrt{x-3} \\\\ (1)^2=(\sqrt{x-3})^2 \\\\ 1=x-3 \\\\ 1+3=x \\\\ 4=x \\\\ x=4 .\end{array}
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