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 - 9-4 Factoring to Solve Quadratic Equations - Got It? - Page 569: 3

Answer

a) $x = -7$ b) The short answer to this is that it is a square function, and square functions have one repeated solution. The long answer to this is that the discriminant is 0. How? $D = b^2 - 4ac$ $D = 4a^2x^2 - 4a^2x^2$ $D = 0$ When the discriminant is zero, there is one repeated root.

Work Step by Step

a) Convert the equation to standard form $x^2 + 14x +49 = 0$ Now, use factoring $(x+7)(x+7) = 0$ Use the zero-product property $x+7 = 0$ or $x+7 = 0$ $x = -7$ or $x = -7$ The solution is -7. b) The short answer to this is that it is a square function, and square functions have one repeated solution. The long answer to this is that the discriminant is 0. How? $D = b^2 - 4ac$ $D = 4a^2x^2 - 4a^2x^2$ $D = 0$ When the discriminant is zero, there is one repeated root.
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