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 - Page 164: 158

Answer

$$0 = x^2 + 6x + 8$$ $$becomes$$ $$x=\frac{-(6)\frac{+}{} \sqrt{(6)^2-4(1)(8)}}{2(1)}$$

Work Step by Step

The quadratic formula $x=\frac{-b\frac{+}{} \sqrt{b^2-4ac}}{2a}$ merely requires direct substitution of the coefficients of any quadratic equation in the form $0 = ax^2 + bx + c$. For the case of $0 = x^2 + 6x + 8$: $$x=\frac{-(6)\frac{+}{} \sqrt{(6)^2-4(1)(8)}}{2(1)}$$ $$x=\frac{-6\frac{+}{} \sqrt{36-32}}{2}$$ $$x=\frac{-6\frac{+}{} \sqrt{4}}{2}$$ $$x=\frac{-6\frac{+}{} 2}{2}$$ $$x = -4$$ $$or$$ $$x = -2$$
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