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: 50

Answer

The solution set is {$\frac{-1-{i}\sqrt 15}{8},\frac{-1+{i}\sqrt 15}{8}$}. No real solution.

Work Step by Step

$4t^2+t+1=0$ $a=4, b=1, c=1$ $D=b^2-4ac=(1)^2-4\cdot4(1)=1-16=-15<0$ The equation has two complex solutions. $x=\frac{-b\pm\sqrt {b^2-4ac}}{2a}=\frac{-1\pm\sqrt -15}{2\cdot4}$ Use the quadratic formula. $x=\frac{-1\pm{i}\sqrt 15}{8}$ $x=\frac{-1+{i}\sqrt 15}{8}$,$x=\frac{-1-{i}\sqrt 15}{8}$ The solution set is {$\frac{-1-{i}\sqrt 15}{8},\frac{-1+{i}\sqrt 15}{8}$}. No real solution.
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