## Elementary and Intermediate Algebra: Concepts & Applications (6th Edition)

Published by Pearson

# Chapter 11 - Quadratic Functions and Equations - 11.2 The Quadratic Formula - 11.2 Exercise Set - Page 713: 33

#### Answer

$p=24$ or $x=\frac{3}{2}$

#### Work Step by Step

$51p=2p^{2}+72\qquad$...add $(-2p^{2}-72)$ to both sides. $51p-2p^{2}-72=2p^{2}+72-2p^{2}-72\qquad$...add like terms. $-2p^{2}+51p-72=0\qquad$...mulitiply both sides by $-1$. $2p^{2}-51p+72=0\qquad$... solve with the Quadractic formula. $a=2,\ b=-51,\ c=72$ $p=\displaystyle \frac{-b\pm\sqrt{b^{2}-4ac}}{2a}\qquad$... substitute $b$ for $-51,\ a$ for $2$ and $c$ for $72$. $p=\displaystyle \frac{-(-51)\pm\sqrt{(-51)^{2}-4\cdot(72)\cdot 2}}{2\cdot 2}\qquad$... simplify. $p=\displaystyle \frac{51\pm\sqrt{2601-576}}{4}$ $p=\displaystyle \frac{51\pm\sqrt{2025}}{4}$ $p=\displaystyle \frac{51\pm 45}{4}\qquad$... the symbol $\pm$ indicates two solutions. $p=\displaystyle \frac{51+45}{4}=\frac{96}{4}=24$ or $x=\displaystyle \frac{51-45}{4}=\frac{6}{4}=\frac{3}{2}$

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