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 - 11.2 The Quadratic Formula - 11.2 Exercise Set - Page 713: 31


$x=10$ or $x=5$

Work Step by Step

$ 14(x-4)-(x+2)=(x+2)(x-4)\qquad$... use the distributive property: $a(b+c)=ab+ac$. $ 14x-56-x-2=(x+2)(x-4)\qquad$... use the FOIL method. $ 14x-56-x-2=x^{2}-4x+2x-8\qquad$...add like terms. $ 13x-58=x^{2}-2x-8\qquad$...add $(-x^{2}+2x+8)$ to both sides. $ 13x-58-x^{2}+2x+8=x^{2}-2x-8-x^{2}+2x+8\qquad$...add like terms. $-x^{2}+15x-50=0\qquad$...mulitiply both sides by $-1$. $ x^{2}-15x+50=0\qquad$... solve with the Quadractic formula. $a=1,\ b=-15,\ c=50$ $ x=\displaystyle \frac{-b\pm\sqrt{b^{2}-4ac}}{2a}\qquad$... substitute $b$ for $-15,\ a$ for $1$ and $c$ for $50$. $ x=\displaystyle \frac{-(-15)\pm\sqrt{(-15)^{2}-4\cdot(50)\cdot 1}}{2\cdot 1}\qquad$... simplify. $x=\displaystyle \frac{15\pm\sqrt{225-200}}{2}$ $x=\displaystyle \frac{15\pm\sqrt{25}}{2}$ $ x=\displaystyle \frac{15\pm 5}{2}\qquad$... the symbol $\pm$ indicates two solutions. $x=\displaystyle \frac{15+5}{2}=\frac{20}{2}=10$ or $x=\displaystyle \frac{15-5}{2}=\frac{10}{2}=5$
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