College Algebra (10th Edition)

Published by Pearson
ISBN 10: 0321979478
ISBN 13: 978-0-32197-947-6

Chapter 1 - Section 1.4 - Radical Equations; Equations Quadratic in Form; Factorable Equations - 1.4 Assess Your Understanding - Page 118: 51


$\displaystyle x=-\frac{1}{3}$

Work Step by Step

We solve by factoring: $(3x+4)^{2}-6(3x+4)+9=0$ $[(3x+4)-3][(3x+4)-3]=0$ $(3x+1)^2=0$ $3x+1=0$ $x=-\frac{1}{3}$
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