Intermediate Algebra (12th Edition)

Published by Pearson
ISBN 10: 0321969359
ISBN 13: 978-0-32196-935-4

Chapter 7 - Section 7.6 - Solving Equations with Radicals - 7.6 Exercises: 32


no solution

Work Step by Step

$\bf{\text{Solution Outline:}}$ To solve the given radical equation, $ \sqrt{x^2-15x+15}=x-5 ,$ square both sides of the equation. Use special products and concepts of solving quadratic equations. Finally, do checking of the solution with the original equation. $\bf{\text{Solution Details:}}$ Squaring both sides of the equation results to \begin{array}{l}\require{cancel} \left( \sqrt{x^2-15x+15} \right)^2=(x-5)^2 \\\\ x^2-15x+15=(x-5)^2 .\end{array} 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 expression above is equivalent to \begin{array}{l}\require{cancel} x^2-15x+15=(x)^2-2(x)(5)+(5)^2 \\\\ x^2-15x+15=x^2-10x+25 .\end{array} Using the properties of equality to isolate the variable results to \begin{array}{l}\require{cancel} (x^2-x^2)+(-15x+10x)=25-15 \\\\ -5x=10 \\\\ x=\dfrac{10}{-5} \\\\ x=-2 .\end{array} Upon checking, $ x=-2 $ DOES NOT satisfy the original equation. Hence, there is $\text{ no solution .}$
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