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: 67c



Work Step by Step

The equation from part b): $600+464=(20+x)(30+x)\\1064=x^2+50x+600\\x^2+50x-464=0\\(x+58)(x-8)=0$ Thus $x=8$ or $x=−58$ but $x$ has to be positive, thus $x=8$.
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