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: 63


The x-intercepts are $(5+\sqrt{47},0)$ and $(-5-\sqrt{47},0)$

Work Step by Step

$ f(x)=x^{2}-10x-22\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}-10x-22\qquad$..add $22$ to both sides so we can complete the square. $ 22=x^{2}-10x\qquad$..add $25$ to both sides to complete the square ($\displaystyle \frac{1}{2}(-10)=-5$, and $(-5)^{2}=25$.) $ x^{2}-10x+25=22+25\qquad$...simplify by applying the Perfect square formula ($(x+a)^{2}=x^{2}+2ax+a^{2}$) and adding like terms. $(x-5)^{2}=47$ 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-5=\pm\sqrt{47}\qquad$...add $5$ to each side. $ x-5+5=\pm\sqrt{47}+5\qquad$...simplify. $x=5\pm\sqrt{47}$ $x=5+\sqrt{47}$ or $x=-5-\sqrt{47}$
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