College Algebra (11th Edition)

Published by Pearson
ISBN 10: 0321671791
ISBN 13: 978-0-32167-179-0

Chapter 1 - Section 1.6 - Other Types of Equations and Applications - 1.6 Exercises - Page 135: 39



Work Step by Step

$\bf{\text{Solution Outline:}}$ To solve the given radical equation, $ \sqrt{4x+5}-6=2x-11 ,$ use the properties of equality to isolate the radical and then square both sides. Then use the properties of equality to express the resulting equation in the form $ax^2+bx+c=0.$ Next is to factor the quadratic equation and solve for the value/s of the variable using the Zero Product Property. Finally, it is a must to do checking of the solution. $\bf{\text{Solution Details:}}$ Using the properties of equality to isolate the radical, the equation above is equivalent to \begin{array}{l}\require{cancel} \sqrt{4x+5}=2x-11+6 \\\\ \sqrt{4x+5}=2x-5 .\end{array} Squaring both sides and then using the special product on squaring binomials 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} 4x+5=(2x-5)^2 \\\\ 4x+5=[(2x)^2-2(2x)(5)+(5)^2] \\\\ 4x+5=4x^2-20x+25 \\\\ -4x^2+(4x+20x)+(5-25)=0 \\\\ -4x^2+24x-20=0 \\\\ \dfrac{-4x^2+24x-20}{-4}=\dfrac{0}{-4} \\\\ x^2-6x+5=0 .\end{array} In the trinomial expression above, the value of $ac$ is $ 1(5)=5 $ and the value of $b$ is $ -6 .$ The $2$ numbers that have a product of $ac$ and a sum of $b$ are $\left\{ -1,-5 \right\}.$ Using these $2$ numbers to decompose the middle term of the trinomial expression above results to \begin{array}{l}\require{cancel} x^2-x-5x+5=0 .\end{array} Grouping the first and second terms and the third and fourth terms, the given expression is equivalent to \begin{array}{l}\require{cancel} (x^2-x)-(5x-5)=0 .\end{array} Factoring the $GCF$ in each group results to \begin{array}{l}\require{cancel} x(x-1)-5(x-1)=0 .\end{array} Factoring the $GCF= (x-1) $ of the entire expression above results to \begin{array}{l}\require{cancel} (x-1)(x-5)=0 .\end{array} Equating each factor to zero (Zero Product Property), then \begin{array}{l}\require{cancel} x-1=0 \\\\\text{OR}\\\\ x-5=0 .\end{array} Solving each equation results to \begin{array}{l}\require{cancel} x-1=0 \\\\ x=1 \\\\\text{OR}\\\\ x-5=0 \\\\ x=5 .\end{array} Upon checking, only $ x=5 $ satisfies the original equation.
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