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: 67

Answer

$\left\{-7, 1\right\}$

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: $x^2+6x+9=7+9 \\x^2+6x+9=16 \\(x+3)^2=16$ Take the square root of both sides to obtain: $x+3=\pm \sqrt{16} \\x+3 = \pm\sqrt{4^2} \\x+3=\pm4$ Subtract 3 on both sides of the equation to obtain: $x = \pm4 - 3$ $x = 4-3 = 1$ or $x = -4-3 = -7$ Therefore, the solution of the given equation is $\left\{-7, 1\right\}$.
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