College Algebra (6th Edition)

Published by Pearson
ISBN 10: 0-32178-228-3
ISBN 13: 978-0-32178-228-1

Chapter 1 - Equations and Inequalities - Exercise Set 1.5 - Page 161: 102


The solutions are $x=\dfrac{1}{5}\pm\dfrac{\sqrt{14}}{5}i$

Work Step by Step

$5x^{2}=2x-3$ Take all terms to the right side of the equation: $5x^{2}-2x+3=0$ Use the quadratic formula to solve this equation. The formula is $x=\dfrac{-b\pm\sqrt{b^{2}-4ac}}{2a}$ In this case, $a=5$, $b=-2$ and $c=3$ Substitute the known values into the formula and evaluate: $x=\dfrac{-(-2)\pm\sqrt{(-2)^{2}-4(5)(3)}}{2(5)}=\dfrac{2\pm\sqrt{4-60}}{10}=...$ $...=\dfrac{2\pm\sqrt{-56}}{10}=\dfrac{2\pm2\sqrt{14}i}{10}=\dfrac{1}{5}\pm\dfrac{\sqrt{14}}{5}i$
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