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: 57


$x=\frac{-3-\sqrt {13}}{2},\frac{-3+\sqrt {13}}{2}$

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 $x$. $x^2+3x-1=0$ $(\frac{3}{2})^2$ $x^2+3x-1+1+(\frac{3}{2})^2=1+(\frac{3}{2})^2$ $x^2+3x+(\frac{3}{2})^2=1+(\frac{3}{2})^2$ $x^2+3x+\frac{9}{4}=1+\frac{9}{4}$ $x^2+3x+\frac{9}{4}=\frac{13}{4}$ $(x+\frac{3}{2})^2=\frac{13}{4}$ $x+\frac{3}{2}=±\frac{\sqrt {13}}{2}$ $x=-\frac{3}{2}±\frac{\sqrt {13}}{2}$ $x=\frac{-3-\sqrt {13}}{2},\frac{-3+\sqrt {13}}{2}$
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