Algebra 1: Common Core (15th Edition)

Published by Prentice Hall
ISBN 10: 0133281140
ISBN 13: 978-0-13328-114-9

Chapter 10 - Radical Expressions and Equations - 10-1 The Pythagorean Theorem - Practice and Problem-Solving Exercises - Page 618: 42


a = 3

Work Step by Step

Notice that a, a+1, and a+2 are the consecutive integers, with a+2 being the greatest. Therefore: $a^{2}+(a+1)^{2}=(a+2)^{2}$ Next, we expand the equation above; $2a^{2}+2a+1=a^{2}+4a+4$ Then, we simplify: $a^{2}-2a-3=0$ Factoring the equation, we find that: $(a-3)(a+1)$ Therefore; $a=3$ (because it cant be negative)
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