Precalculus (6th Edition) Blitzer

Published by Pearson
ISBN 10: 0-13446-914-3
ISBN 13: 978-0-13446-914-0

Chapter P - Section P.7 - Equations - Exercise Set: 68

Answer

$\left\{-4, -2\right\}$

Work Step by Step

RECALL: To solve $x^2 + bx=c$ by completing the square, add $\left(\frac{b}{2}\right)^2$ on both sides of the equation. Thus, to solve the given equation by completing the square, add $(\frac{6}{2})^2=3^2=9$ on both sides of the equation to obtain: $x^2+6x+9=-8+9 \\x^2+6x+9=1 \\(x+3)^2=1$ Take the square root of both sides to obtain: $x+3=\pm \sqrt{1} \\x+3 = \pm1$ Subtract 3 on both sides of the equation to obtain: $x = \pm1 - 3 \\x_1 = 1-3 = -2 \\x_2 = -1-3 = -4$ Therefore, the solution of the given equation is: $\left\{-4, -2\right\}$.
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