Algebra 2 (1st Edition)

Published by McDougal Littell
ISBN 10: 0618595414
ISBN 13: 978-0-61859-541-9

Chapter 4 Quadratic Functions and Factoring - 4.3 Solve x(squared) + bx + c = 0 - 4.3 Exercises - Problem Solving - Page 257: 65


$150$ and $100$

Work Step by Step

Our equation is according to the text of the exercise: $3\cdot100\cdot50=(100+x)(50+x)\\15000=5000+x^2+150x\\x^2+150x-10000=0\\(x+200)(x-50)=0$ Thus $x=50$ or $x=−200$ but $x$ has to be positive, thus $x=50$. Thus the new dimensions are $150$ and $100$.
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