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: 13


$x=\displaystyle \frac{3}{2}+\frac{\sqrt{29}}{2}$ or $x=\displaystyle \frac{3}{2}-\frac{\sqrt{29}}{2}$

Work Step by Step

$ x^{2}=3x+5\qquad$...add $-3x-5$ to both sides. $x^{2}-3x-5=3x+5-3x-5$ $ x^{2}-3x-5=0\qquad$... solve with the Quadractic formula. $a=1,\ b=-3,\ c=-5$ $ x=\displaystyle \frac{-b\pm\sqrt{b^{2}-4ac}}{2a}\qquad$... substitute $b$ for $-3,\ a$ for $1$ and $c$ for $-5$. $ x=\displaystyle \frac{-(-3)\pm\sqrt{(-3)^{2}-4\cdot(-5)\cdot 1}}{2\cdot 1}\qquad$... simplify. $x=\displaystyle \frac{3\pm\sqrt{9+20}}{2}$ $ x=\displaystyle \frac{3\pm\sqrt{29}}{2}\qquad$... the symbol $\pm$ indicates two solutions. $x=\displaystyle \frac{3}{2}+\frac{\sqrt{29}}{2}$ or $x=\displaystyle \frac{3}{2}-\frac{\sqrt{29}}{2}$
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