Introductory Algebra for College Students (7th Edition)

Published by Pearson
ISBN 10: 0-13417-805-X
ISBN 13: 978-0-13417-805-9

Chapter 9 - Section 9.3 - The Quadratic Formula - Exercise Set - Page 647: 2


The solution set is {-2, -5}.

Work Step by Step

The quadratic formula is given by: $$x = \frac{-b ± \sqrt {b^{2} - 4ac}}{2a}$$ To solve a quadratic equation by the quadratic formula, we need to know what $a$, $b$, and $c$ are. The coefficient of the first term is $a$. The coefficient of the second term is $b$. The constant term is $c$. Thus, we see that $a$ is $1$, $b$ is $7$, and $c$ is $10$. Now, we can substitute these values into the equation: $$x = \frac{-7 ± \sqrt {7^{2} - 4(1)(10)}}{2(1)}$$ We simplify exponents first, according to the order of operations: $$x = \frac{-7 ± \sqrt {49 - 4(1)(10)}}{2(1)}$$ Now we simplify what is under the radical sign: $$x = \frac{-7 ± \sqrt {49 - 40}}{2(1)}$$ $$x = \frac{-7 ± \sqrt {9}}{2(1)}$$ Now, we evaluate the radical: $$x = \frac{-7 ± 3}{2(1)}$$ Now, we simplify the numerator, and we end up with two options because of the ± sign: $$x = \frac{-4}{2(1)}$$ and $$x = \frac{-10}{2(1)}$$ Simplify the denominators: $$x = \frac{-4}{2}$$ and $$x = \frac{-10}{2}$$ We simplify the fractions: $$x = -2$$ and $$x = -5$$ The solution set is {-2, -5}.
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