Algebra and Trigonometry 10th Edition

Published by Cengage Learning
ISBN 10: 9781337271172
ISBN 13: 978-1-33727-117-2

Chapter 1 - 1.4 - Quadratic Equations and Applications - 1.4 Exercises - Page 111: 89



Work Step by Step

We are given the equation: $\dfrac{1}{2}x^2+\dfrac{3}{8}x=2$ Multiply all terms by 8 to clear denominators, then write the equation in standard form: $8\cdot\dfrac{1}{2}x^2+8\cdot\dfrac{3}{8}x=8\cdot2$ $4x^2+3x=16$ $4x^2+3x-16=0$ Solve the equation using the quadratic formula: $x=\dfrac{-3\pm\sqrt{3^2-4(4)(-16)}}{2(4)}=\dfrac{-3\pm\sqrt{265}}{8}$ $x_1=\dfrac{-3-\sqrt{265}}{8}$ $x_2=\dfrac{-3+\sqrt{265}}{8}$ The solution set of the equation is: $\left\{\dfrac{-3-\sqrt{265}}{8},\dfrac{-3+\sqrt{265}}{8}\right\}$
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