College Algebra 7th Edition

Published by Brooks Cole
ISBN 10: 1305115546
ISBN 13: 978-1-30511-554-5

Chapter 1, Equations and Graphs - Section 1.4 - Solving Quadratic Equations - 1.4 Exercises: 35

Answer

$x=-1\displaystyle \pm\frac{2\sqrt{6}}{3}$

Work Step by Step

We solve by completing the square: $3x^{2}+6x-5=0$ $x^{2}+2x-\frac{5}{3}=0$ $x^{2}+2x=\frac{5}{3}$ $x^{2}+2x+1=\frac{5}{3}+1$ $(x+1)^{2}=\frac{8}{3}$ $x+1=\pm\sqrt{\frac{8}{3}}=\pm\sqrt{\frac{4*2*3}{3*3}}=\pm\frac{2\sqrt{6}}{3}$ $x=-1\displaystyle \pm\frac{2\sqrt{6}}{3}$
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