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: 62

Answer

$x=\left\{ \dfrac{1-\sqrt{13}}{2},\dfrac{1+\sqrt{13}}{2} \right\}$

Work Step by Step

Using the properties of equality, the given equation, $ 0.1x^2-0.1x=0.3 ,$ is equivalent to \begin{array}{l}\require{cancel} 10(0.1x^2-0.1x)=(0.3)10 \\\\ 1x^2-1x=3 \\\\ x^2-x=3 \\\\ x^2-x-3=0 .\end{array} The equation above has $a=1$, $b=-1$, 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{-(-1)\pm\sqrt{(-1)^2-4(1)(-3)}}{2(1)} \\\\ x=\dfrac{1\pm\sqrt{1+12}}{2} \\\\ x=\dfrac{1\pm\sqrt{13}}{2} .\end{array} Hence, the solutions are $ x=\left\{ \dfrac{1-\sqrt{13}}{2},\dfrac{1+\sqrt{13}}{2} \right\} .$
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