Elementary and Intermediate Algebra: Concepts & Applications (6th Edition)

Published by Pearson
ISBN 10: 0-32184-874-8
ISBN 13: 978-0-32184-874-1

Chapter 11 - Quadratic Functions and Equations - 11.1 Quadratic Equations - 11.1 Exercise Set - Page 706: 62

Answer

$\left(-\frac{5}{2}-\frac{\sqrt{17}}{2}, 0\right)$ and $\left(-\frac{5}{2}+\frac{\sqrt{17}}{2}, 0\right)$

Work Step by Step

The $x$-intercepts of a function can be found by setting $y=0$ then solving for $x$. Replace $g(x)$ by $0$, then solve for $x$ to obtain: \begin{align*} 0&=x^2+5x+2\\ -25&=x^2+5x \end{align*} Complete the square by adding $\left(\dfrac{5}{2}\right)^2=\dfrac{25}{4}$ to both sides to obtain: \begin{align*} -2+\frac{25}{4}&=x^2+5x+\frac{25}{4}\\\\ -\frac{8}{4}+\frac{25}{4}&=\left(x+\frac{5}{2}\right)^2\\\\ \frac{17}{4}&=\left(x+\frac{5}{2}\right)^2\\\\ \pm\sqrt{\frac{17}{4}}&=\sqrt{\left(x+\frac{5}{2}\right)^2}\\\\ \pm\frac{\sqrt{17}}{2}&=x+\frac{5}{2}\\\\ -\frac{5}{2}\pm\frac{\sqrt{17}}{2}&=x \end{align*} Thus, the $x$-intercepts are: $\left(-\frac{5}{2}-\frac{\sqrt{17}}{2}, 0\right)$ and $\left(-\frac{5}{2}+\frac{\sqrt{17}}{2}, 0\right)$
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