College Algebra (11th Edition)

Published by Pearson
ISBN 10: 0321671791
ISBN 13: 978-0-32167-179-0

Chapter 1 - Section 1.4 - Quadratic Equations - 1.4 Exercises - Page 112: 61

Answer

$x=\left\{ \dfrac{-2-\sqrt{10}}{2},\dfrac{-2+\sqrt{10}}{2} \right\} $

Work Step by Step

Using the properties of equality, the given equation, $ 0.2x^2+0.4x-0.3=0 ,$ is equivalent to \begin{array}{l}\require{cancel} 10\left(0.2x^2+0.4x-0.3\right)=(0)10 \\\\ 2x^2+4x-3=0 .\end{array} The equation above has $a=2$, $b=4$, and $c=-3$ Using $x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}$ or the Quadratic Formula, the solutions to the equation above are \begin{array}{l}\require{cancel} x=\dfrac{-4\pm\sqrt{4^2-4(2)(-3)}}{2(2)} \\\\ x=\dfrac{-4\pm\sqrt{16+24}}{4} \\\\ x=\dfrac{-4\pm\sqrt{40}}{4} \\\\ x=\dfrac{-4\pm\sqrt{4\cdot10}}{4} \\\\ x=\dfrac{-4\pm\sqrt{(2)^2\cdot10}}{4} \\\\ x=\dfrac{-4\pm\sqrt{(2)^2\cdot10}}{4} \\\\ x=\dfrac{-4\pm2\sqrt{10}}{4} \\\\ x=\dfrac{2(-2\pm\sqrt{10})}{4} \\\\ x=\dfrac{\cancel{2}(-2\pm\sqrt{10})}{\cancel{2}(2)} \\\\ x=\dfrac{-2\pm\sqrt{10}}{2} .\end{array} Hence, the solutions are $ x=\left\{ \dfrac{-2-\sqrt{10}}{2},\dfrac{-2+\sqrt{10}}{2} \right\} .$
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