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: 64


The x-intercepts are $(4+\sqrt{26},0)$ and $(4-\sqrt{26},0)$

Work Step by Step

$ f(x)=x^{2}-8x-10\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}-8x-10\qquad$..add $10$ to both sides so we can complete the square. $ 10=x^{2}-8x\qquad$..add $16$ to both sides to complete the square ($\displaystyle \frac{1}{2}(-8)=-4$, and $(-4)^{2}=16$.) $ x^{2}-8x+16=10+16\qquad$...simplify by applying the Perfect square formula ($(x+a)^{2}=x^{2}+2ax+a^{2}$) and adding like terms. $(x-4)^{2}=26$ 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-4=\pm\sqrt{26}\qquad$...add $4$ to each side. $ x-4+4=\pm\sqrt{26}+4\qquad$...simplify. $x=4\pm\sqrt{26}$ $x=4+\sqrt{26}$ or $x=4-\sqrt{26}$
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