College Algebra (10th Edition)

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

Chapter 1 - Section 1.2 - Quadratic Equations - 1.2 Assess Your Understanding - Page 101: 53


The solution set is {$0,\frac{9}{4}$}.

Work Step by Step

$4x^2-9x=0$ $a=4, b=-9, c=0$ $D=b^2-4ac=(-9)^2-4\cdot4(0)=81-0=81>0$ The equation has two real solutions. $x=\frac{-b\pm\sqrt {b^2-4ac}}{2a}=\frac{9\pm\sqrt 81}{2\cdot4}$ Use the quadratic formula. $x=\frac{9\pm9}{8}$ $x=\frac{18}{8}$ , $x=\frac{9-9}{8}$ $x=\frac{9}{4}$,$x=0$ The solution set is {$0,\frac{9}{4}$}.
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