Intermediate Algebra (6th Edition)

Published by Pearson
ISBN 10: 0321785045
ISBN 13: 978-0-32178-504-6

Chapter 5 - Section 5.8 - Solving Equations by Factoring and Problem Solving - Exercise Set - Page 323: 77


$2$ inches

Work Step by Step

The area of the rectangular board and the border is $A_1=(2x+12)(2x+16),$ while the area of the rectangular board is $A_2=12(16).$ Subtracting the two areas results to the area of the border. Since, the area of the border is given as $128$ square inches, then \begin{array}{l}\require{cancel} A=A_1-A_2 \\ 128=(2x+12)(2x+16)-12(16) \\ 128=(2x+12)(2x+16)-192 \\ 128+192=(2x+12)(2x+16) \\ 320=(2x+12)(2x+16) .\end{array} Using the FOIL Method which is given by $(a+b)(c+d)=ac+ad+bc+bd,$ the expression above is equivalent to \begin{array}{l}\require{cancel} 320=2x(2x)+2x(16)+12(2x)+12(16) \\ 320=4x^2+32x+24x+192 \\ 0=4x^2+32x+24x+192-320 \\ 4x^2+56x-128=0 \\ \dfrac{4x^2+56x-128}{4}=\dfrac{0}{4} \\ x^2+14x-32=0 \\ (x+16)(x-2)=0 .\end{array} Equating each factor to zero (Zero Product Property) then the solutions of the equation above are $ x=\{-16,2\} .$ Since $x$ is a measurement, then only $x=2$ is the acceptable solution. Therefore, the width of the border, $x,$ is $2$ inches.
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