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 - Connecting: The Concepts - Exercises - Page 712: 3

Answer

$x=-3+\sqrt{19}$ or $x=-3-\sqrt{19}$

Work Step by Step

$x^{2}+6x=10$ $\qquad$...square half the coefficient of $x$ and add it to both sides to complete the square $(\displaystyle \frac{6}{2})=3,\ 3^{2}=9$ $ x^{2}+6x+9=19\qquad$...write $x^{2}+6x+9$ as a binomial squared. $(x+3)^{2}=19\qquad$...use 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{19}\qquad$...add $-3$ to both sides. $ x=-3\pm\sqrt{19}\qquad$... the symbol $\pm$ indicates two solutions. $x=-3+\sqrt{19}$ or $x=-3-\sqrt{19}$
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