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: 61


The solutions are $x=-1\pm\dfrac{\sqrt{10}}{2}$

Work Step by Step

$0.2x^{2}+0.4x-0.3=0$ Multiply the whole equation by $10$: $10(0.2x^{2}+0.4x-0.3=0)$ $2x^{2}+4x-3=0$ Use the quadratic formula to solve this equation. The formula is $x=\dfrac{-b\pm\sqrt{b^{2}-4ac}}{2a}$. In this case, $a=2$, $b=4$ and $c=-3$ Substitute the known values into the formula and evaluate: $x=\dfrac{-4\pm\sqrt{4^{2}-4(2)(-3)}}{2(2)}=\dfrac{-4\pm\sqrt{16+24}}{4}=...$ $...=\dfrac{-4\pm\sqrt{40}}{4}=\dfrac{-4\pm2\sqrt{10}}{4}=-\dfrac{4}{4}\pm\dfrac{2\sqrt{10}}{4}=...$ $...=-1\pm\dfrac{\sqrt{10}}{2}$ The solutions are $x=-1\pm\dfrac{\sqrt{10}}{2}$
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