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 102: 81


The solution set is $\left\{\frac{1}{4}\right\}$.

Work Step by Step

RECALL: A quadratic equation of the form $ax^2+bx+c=0$ can be solved using the quadratic formula $x=\dfrac{-b \pm \sqrt{b^2-4ac}}{2a}$ The given quadratic equation has: $a=16 \\b=-8 \\c=1$ Substitute these values into the quadratic formula to obtain: $x=\dfrac{-(-8) \pm \sqrt{(-8)^2-4(16)(1)}}{2(16)} \\x=\dfrac{8 \pm \sqrt{64-64}}{32} \\x=\dfrac{8\pm \sqrt{0}}{32} \\x =\dfrac{8}{32} \\x=\dfrac{1}{4}$ Therefore, the solution set is $\left\{\frac{1}{4}\right\}$.
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