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: 33

Answer

$r=1$

Work Step by Step

$\bf{\text{Solution Outline:}}$ To solve the given radical equation, $ \sqrt{r^2+9r+15}-r-4=0 ,$ use the properties of equality to isolate the radical expression. Then 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. $\bf{\text{Solution Details:}}$ Using the properties of equality to isolate the radical expression results to \begin{array}{l}\require{cancel} \sqrt{r^2+9r+15}=r+4 .\end{array} Squaring both sides of the equation results to \begin{array}{l}\require{cancel} \left( \sqrt{r^2+9r+15} \right)^2=(r+4)^2 \\\\ r^2+9r+15=(r+4)^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} r^2+9r+15=(r)^2+2(r)(4)+(4)^2 \\\\ r^2+9r+15=r^2+8r+16 .\end{array} Using the properties of equality to isolate the variable results to \begin{array}{l}\require{cancel} (r^2-r^2)+(9r-8r)=16-15 \\\\ r=1 .\end{array} Upon checking, $ r=1 $ satisfies the original equation.
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