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: 83


The solution set is $\left\{-\frac{3}{5}, \frac{5}{2}\right\}$.

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: $a=10 \\b=-19 \\c=-15$ Substitute these values into the quadratic formula to obtain: $x=\dfrac{-(-19) \pm \sqrt{(-19)^2-4(10)(-15)}}{2(10)} \\x=\dfrac{19 \pm \sqrt{361-(-600)}}{20} \\x=\dfrac{19\pm \sqrt{361+600}}{20} \\x =\dfrac{19\pm\sqrt{961}}{20} \\x=\dfrac{19\pm\sqrt{31^2}}{20} \\x=\dfrac{19\pm31}{20}$ Split the solutions to obtain: $x_1 = \dfrac{19+31}{20} = \dfrac{50}{20} = \dfrac{5}{2} \\x_2=\dfrac{19-31}{20}=\dfrac{-12}{20}=-\dfrac{3}{5}$ Therefore, the solution set is $\left\{-\frac{3}{5}, \frac{5}{2}\right\}$.
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