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 - Skill Practice - Page 256: 43



Work Step by Step

Our equation is according to the text of the exercise: $3\cdot12\cdot10=(12+x)(10+x)\\360=120+x^2+22x\\x^2+22x-240=0\\(x-8)(x+30)=0$ Thus $x=8$ or $x=-30$ but $x$ has to be positive, thus $x=8$.
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