College Algebra (6th Edition)

Published by Pearson
ISBN 10: 0-32178-228-3
ISBN 13: 978-0-32178-228-1

Chapter 1 - Equations and Inequalities - Exercise Set 1.5: 157

Answer

$$0 = x^2 + 6x + 8$$ $$becomes$$ $$0 = (x+3)^2 -1$$

Work Step by Step

To complete the square, we use the formula $(\frac{b}{2a})^2$ in any quadratic equation $0= ax^2 + bx + c$. In the case of the exercise $0 = x^2 + 6x + 8$: $$(\frac{(6)}{2(1)})^2 = (3)^2 = 9$$ This means that a perfect square would require a $9$ instead of an $8$ like this exercise has. To complete the square, therefore, we must do as follows: $$0 = x^2 + 6x + 8 + 1 - 1$$ $$0 = (x^2 + 6x +9) - 1$$ $$0 = (x+3)^2 - 1$$ $$1 = (x+3)^2$$ $$\sqrt{1} = \sqrt{(x+3)^2}$$ $$\frac{+}{}1 = x + 3$$ $$x = -2$$ $$or$$ $$x = -4$$
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