Precalculus: Mathematics for Calculus, 7th Edition

Published by Brooks Cole
ISBN 10: 1305071751
ISBN 13: 978-1-30507-175-9

Chapter 1 - Section 1.6 - Complex Numbers - 1.6 Exercises - Page 64: 71



Work Step by Step

$6x^{2}+12x+7=0$ Use the quadratic formula to solve this equation. The formula is $x=\dfrac{-b\pm\sqrt{b^{2}-4ac}}{2a}$. For this equation, $a=6$, $b=12$ and $c=7$. Substitute the known values in the formula: $x=\dfrac{-12\pm\sqrt{12^{2}-4(6)(7)}}{2(6)}=\dfrac{-12\pm\sqrt{144-168}}{122}=...$ $...=\dfrac{-12\pm\sqrt{-24}}{12}=\dfrac{-12\pm2\sqrt{6}i}{12}=-1\pm\dfrac{\sqrt{6}}{6}i$ The answer is $x=-1\pm\dfrac{\sqrt{6}}{6}i$
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