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: 38

Answer

$x=5$

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.
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.