Precalculus (6th Edition)

Published by Pearson
ISBN 10: 013421742X
ISBN 13: 978-0-13421-742-0

Chapter 1 - Equations and Inequalities - 1.4 Quadratic Equations - 1.4 Exercises - Page 122: 62


The solutions are $x=\dfrac{1}{2}\pm\dfrac{\sqrt{13}}{2}$

Work Step by Step

$0.1x^{2}-0.1x=0.3$ Multiply the whole equation by $10$: $10(0.1x^{2}-0.1x=0.3)$ $x^{2}-x=3$ Take $3$ to the left side: $x^{2}-x-3=0$ Use the quadratic formula to solve this equation. The formula is $x=\dfrac{-b\pm\sqrt{b^{2}-4ac}}{2a}$. In this case, $a=1$, $b=-1$ and $c=−3$ Substitute the known values into the formula and evaluate: $x=\dfrac{-(-1)\pm\sqrt{(-1)^{2}-4(1)(-3)}}{2(1)}=\dfrac{1\pm\sqrt{1+12}}{2}=...$ $...=\dfrac{1\pm\sqrt{13}}{2}=\dfrac{1}{2}\pm\dfrac{\sqrt{13}}{2}$ The solutions are $x=\dfrac{1}{2}\pm\dfrac{\sqrt{13}}{2}$
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