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: 51

Answer

The solution set is {$\frac{-1-\sqrt 5}{4},\frac{-1+\sqrt 5}{4}$}.

Work Step by Step

$4x^2+2x-1=0$ $a=4, b=2, c=-1$ $D=b^2-4ac=(2)^2-4\cdot4(-1)=4+16=20>0$ The equation has two real solutions. $x=\frac{-b\pm\sqrt {b^2-4ac}}{2a}=\frac{-2\pm\sqrt 20}{2\cdot4}$ Use the quadratic formula. $x=\frac{-2\pm2\sqrt 5}{8}$ $x=\frac{-1+\sqrt 5}{4}$,$x=\frac{-1-\sqrt 5}{4}$ The solution set is {$\frac{-1-\sqrt 5}{4},\frac{-1+\sqrt 5}{4}$}.
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