Precalculus (6th Edition)

Published by Pearson
ISBN 10: 013421742X
ISBN 13: 978-0-13421-742-0

Chapter 1 - Equations and Inequalities - 1.5 Applications and Modeling with Quadratic Equations - 1.5 Exercises - Page 130: 23


100 and 400

Work Step by Step

Let the width of parking lot $= x$ Length of parking lot $= 2x+200$ Area $= 40,000$ Area of rectangle = $length\times breadth$ $(2x+200)x=40,000$ Open the brackets, $2x^{2}+200x=40,000$ Divide the entire equation by 2, $x^{2}+100x=20,000$ Shift the 20,000 to Left Side. $x^{2}+100x-20,000=0$ Solve the quadratic by factorizing $(x+200)(x-100)=0$ Thus, $x= -200 , 100$ Discard the negative value. (length cannot be negative) Therefore width $= 100$ Length $= 2(100)+200= 400$
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