College Algebra (10th Edition)

Published by Pearson
ISBN 10: 0321979478
ISBN 13: 978-0-32197-947-6

Chapter 1 - Section 1.2 - Quadratic Equations - 1.2 Assess Your Understanding - Page 102: 69


$x=\frac{-\sqrt{3}\pm\sqrt{15}}{2}$ $x=1.07$ or $x=-2.80$

Work Step by Step

We are given: $x^{2}+\sqrt{3}x-3=0$ We solve using the quadratic formula ($a=1,\ b=\sqrt{3},\ c=-3$): $x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$ $x=\frac{-\sqrt{3}\pm\sqrt{(\sqrt{3})^{2}-4*1*-3}}{2*1}$ $x=\frac{-\sqrt{3}\pm\sqrt{3+12}}{2}$ $x=\frac{-\sqrt{3}\pm\sqrt{15}}{2}$ $x=1.07$ or $x=-2.80$
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