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: 59


The solutions are $x=-3+\sqrt{2}$ or $x=-3-\sqrt{2}$

Work Step by Step

$ f(x)=x^{2}+6x+7\qquad$..an x-intercept is a point in the graph where the function intercepts the $\mathrm{x}$ axis. If a function intercepts the $x$ axis, $f(x)=0$. Substitute $0$ for $f(x)$ in the given function. $ 0=x^{2}+6x+7\qquad$..add $-7$ to both sides so we can complete the square on the right side. $-7=x^{2}+6x\qquad$..add $9$ to both sides to complete the square ($\displaystyle \frac{1}{2}(6)=3$, and $(3)^{2}=9$.) $ x^{2}+6x+9=-7+9\qquad$...simplify by applying the Perfect square formula ($(x+a)^{2}=x^{2}+2ax+a^{2}$) and adding like terms. $(x+3)^{2}=2$ According to the general principle of square roots: For any real number $k$ and any algebraic expression $x$ : $\text{If }x^{2}=k,\text{ then }x=\sqrt{k}\text{ or }x=-\sqrt{k}$. $ x+3=\pm\sqrt{2}\qquad$...add $-3$ to each side. $ x+3-3=\pm\sqrt{2}-3\qquad$...simplify. $x=-3\pm\sqrt{2}$ $x=-3+\sqrt{2}$ or $x=-3-\sqrt{2}$
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