College Algebra (6th Edition)

Published by Pearson
ISBN 10: 0-32178-228-3
ISBN 13: 978-0-32178-228-1

Chapter 1 - Equations and Inequalities - Exercise Set 1.5 - Page 160: 58

Answer

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

Work Step by Step

To complete the square, add a term that is the square of half the coefficient of $x$ to both sides of the equation, and factor the left side. Then use the square root property to simplify both sides, and then isolate $x$. $x^2-3x-5=0$ $(\frac{-3}{2})^2$ $x^2-3x-5+5+(\frac{-3}{2})^2=5+(\frac{-3}{2})^2$ $x^2-3x+(\frac{-3}{2})^2=5+(\frac{-3}{2})^2$ $x^2-3x+\frac{9}{4}=5+\frac{9}{4}$ $x^2-3x+\frac{9}{4}=\frac{29}{4}$ $(x-\frac{3}{2})^2=\frac{29}{4}$ $x-\frac{3}{2}=±\frac{\sqrt {29}}{2}$ $x=\frac{3}{2}±\frac{\sqrt {29}}{2}$ $x=\frac{3-\sqrt {29}}{2},\frac{3+\sqrt {29}}{2}$
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