Precalculus: Mathematics for Calculus, 7th Edition

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

Chapter 1 - Review - Exercises - Page 134: 65

Answer

$x=-\dfrac{2}{3}\pm\dfrac{\sqrt{7}}{3}$

Work Step by Step

$3x^{2}+4x-1=0$ Use the quadratic formula to solve this equation. The formula is $x=\dfrac{-b\pm\sqrt{b^{2}-4ac}}{2a}$. For this equation, $a=3$, $b=4$ and $c=-1$. Substitute the known values into the formula and simplify: $x=\dfrac{-4\pm\sqrt{4^{2}-4(3)(-1)}}{2(3)}=\dfrac{-4\pm\sqrt{16+12}}{6}=...$ $...=\dfrac{-4\pm\sqrt{28}}{6}=\dfrac{-4\pm2\sqrt{7}}{6}=-\dfrac{4}{6}\pm\dfrac{2\sqrt{7}}{6}=-\dfrac{2}{3}\pm\dfrac{\sqrt{7}}{3}$
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