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: 34


$p=\frac{4}{3}$ or $x=-18$

Work Step by Step

$ 72=3p^{2}+50p\qquad$...add $(-3p^{2}-50p)$ to both sides. $ 72-3p^{2}-50p=3p^{2}+50p-3p^{2}-50p\qquad$...add like terms. $-3p^{2}-50p+72=0\qquad$...mulitiply both sides by $-1$. $ 3p^{2}+50p-72=0\qquad$... solve with the Quadractic formula. $a=3,\ b=50,\ c=-72$ $ p=\displaystyle \frac{-b\pm\sqrt{b^{2}-4ac}}{2a}\qquad$... substitute $b$ for $50,\ a$ for $3$ and $c$ for $-72$. $ p=\displaystyle \frac{-50\pm\sqrt{(50)^{2}-4\cdot(-72)\cdot 3}}{2\cdot 3}\qquad$... simplify. $p=\displaystyle \frac{-50\pm\sqrt{2500+864}}{6}$ $p=\displaystyle \frac{-50\pm\sqrt{3364}}{6}$ $ p=\displaystyle \frac{-50\pm 58}{6}\qquad$... the symbol $\pm$ indicates two solutions. $p=\displaystyle \frac{-50+58}{6}=\frac{8}{6}=\frac{4}{3}$ or $x=\displaystyle \frac{-50-58}{6}=\frac{-108}{6}=-18$
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