Intermediate Algebra for College Students (7th Edition)

Published by Pearson
ISBN 10: 0-13417-894-7
ISBN 13: 978-0-13417-894-3

Chapter 8 - Section 8.2 - The Quadratic Formula - Concept and Vocabulary Check - Page 607: 11

Answer

$x^2 + 8x + 15 = 0$ can easily be solved by completing the square. Unlike factoring which requires trial and error, using completing the square method will directly give you the solution set by using the square root property. For instance: $x^2 + 8x + 15 = 0$ $x^2 + 8x = -15$ coefficient of the $x$-term = $8$; $\frac{8}{2}=4$; $4^2=16$ Add $16$ to both sides to complete the square: $x^2 + 8x +16= -15+16$ $x^2 + 8x +16= 1$ $(x+4)^2 = 1$ Using the Square Root Property $u^2 = d$, then $u = \sqrt d$ or $u = - \sqrt d$. Thus, $x +4= ±\sqrt{1}$ $x = -4±1$ $x=-3$ $x=-5$

Work Step by Step

$x^2 + 8x + 15 = 0$ can easily be solved by completing the square. Unlike factoring which requires trial and error, using completing the square method will directly give you the solution set by using the square root property. For instance: $x^2 + 8x + 15 = 0$ $x^2 + 8x = -15$ coefficient of the $x$-term = $8$; $\frac{8}{2}=4$; $4^2=16$ Add $16$ to both sides to complete the square: $x^2 + 8x +16= -15+16$ $x^2 + 8x +16= 1$ $(x+4)^2 = 1$ Using the Square Root Property $u^2 = d$, then $u = \sqrt d$ or $u = - \sqrt d$. Thus, $x +4= ±\sqrt{1}$ $x = -4±1$ $x=-3$ $x=-5$
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