Trigonometry (11th Edition) Clone

Published by Pearson
ISBN 10: 978-0-13-421743-7
ISBN 13: 978-0-13421-743-7

Chapter 8 - Quiz (Sections 8.1-8.4) - Page 385: 4



Work Step by Step

Step 1: Comparing $3x^{2}-x+4=0$ to the standard form of a quadratic equation $ax^{2}+bx+c=0$; $a=3$, $b=-1$ and $c=4$ Step 2: The quadratic formula is: $x=\frac{-b \pm \sqrt {b^{2}-4ac}}{2a}$ Step 3: Substituting the values of a,b and c in the formula: $x=\frac{-(-1) \pm \sqrt {(-1)^{2}-4(3)(4)}}{2(3)}$ Step 4: $x=\frac{1 \pm \sqrt {1-48}}{6}$ Step 5: $x=\frac{1 \pm \sqrt {-47}}{6}$ Step 6: $x=\frac{1 \pm \sqrt {-1\times47}}{6}$ Step 7: $x=\frac{1 \pm (\sqrt {-1}\times\sqrt {47})}{6}$ Step 8: $x=\frac{1 \pm (i\times \sqrt {47})}{6}$ Step 9: $x=\frac{1 \pm i\sqrt {47}}{6}$ Step 10: $x=\frac{1}{6}\pm\frac{47i}{6}$ Step 11: Therefore, the solution set is {$\frac{1}{6}\pm\frac{47}{6}i$}.
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