Algebra 1: Common Core (15th Edition)

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

Chapter 9 - Quadratic Functions and Equations - Common Core Cumulative Standards Review - Selected Response - Page 610: 22

Answer

Sometimes. $x^2 + 6x + 8 = 0$ Can be factored to $(x+2)(x+4) = 0$ This has 2 real solutions, -2 and -4 $x^2 + 6x + 9$ Can be factored to $(x+3)(x+3) = 0$ This has 1 real solution, -3

Work Step by Step

Sometimes. $x^2 + 6x + 8 = 0$ Can be factored to $(x+2)(x+4) = 0$ This has 2 real solutions, -2 and -4 $x^2 + 6x + 9$ Can be factored to $(x+3)(x+3) = 0$ This has 1 real solution, -3
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