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 - 10.6 Solving Radical Equations - 10.6 Exercise Set - Page 668: 38



Work Step by Step

Using the properties of equality to isolate the radical expression in the given equation, $ x=\sqrt{x-1}+3 ,$ results to \begin{array}{l}\require{cancel} x=\sqrt{x-1}+3 \\\\ x-3=\sqrt{x-1} .\end{array} Squaring both sides and using the square of a binomial which is given by $(a+b)^2=a^2+2ab+b^2$ or by $(a-b)^2=a^2-2ab+b^2,$ the equation above is equivalent to \begin{array}{l}\require{cancel} x-3=\sqrt{x-1} \\\\ (x-3)^2=\left( \sqrt{x-1} \right)^2 \\\\ (x)^2-2(x)(3)+(3)^2=x-1 \\\\ x^2-6x+9=x-1 \\\\ x^2-6x-x+9+1=0 \\\\ x^2-7x+10=0 .\end{array} Using the FOIL Method which is given by $(a+b)(c+d)=ac+ad+bc+bd,$ the equation above is equivalent to \begin{array}{l}\require{cancel} x^2-7x+10=0 \\\\ (x-5)(x-2)=0 .\end{array} Equating each factor to zero (Zero Product Property, the solutions are \begin{array}{l}\require{cancel} x-5=0 \\\\\text{OR}\\\\ x-2=0 .\end{array} Solving each equation results to \begin{array}{l}\require{cancel} x-5=0 \\\\ x=5 \\\\\text{OR}\\\\ x-2=0 \\\\ x=2 .\end{array} If $x=2,$ then \begin{array}{l}\require{cancel} x=\sqrt{x-1}+3 \\\\ 2=\sqrt{2-1}+3 \\\\ 2=\sqrt{1}+3 \\\\ 2=1+3 \\\\ 2=4 \text{ (FALSE)} .\end{array} Hence, only $ x=5 $ satisfies the original equation.
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