College Algebra (6th Edition)

Published by Pearson
ISBN 10: 0-32178-228-3
ISBN 13: 978-0-32178-228-1

Chapter 1 - Equations and Inequalities - Exercise Set 1.5 - Page 161: 130


The solutions are $x=-\sqrt{3}\pm2i$

Work Step by Step

$\sqrt{3}x^{2}+6x+7\sqrt{3}=0$ Use the quadratic formula to solve this equation. The formula is $x=\dfrac{-b\pm\sqrt{b^{2}-4ac}}{2a}$ In this case, $a=\sqrt{3}$, $b=6$ and $c=7\sqrt{3}$ Substitute the known values into the formula and evaluate: $x=\dfrac{-6\pm\sqrt{6^{2}-4(\sqrt{3})(7\sqrt{3})}}{2\sqrt{3}}=\dfrac{-6\pm\sqrt{36-84}}{2\sqrt{3}}=...$ $...=\dfrac{-6\pm\sqrt{-48}}{2\sqrt{3}}=\dfrac{-6\pm4\sqrt{3}i}{2\sqrt{3}}=-\dfrac{6}{2\sqrt{3}}\pm\dfrac{4\sqrt{3}}{2\sqrt{3}}i=...$ $...=-\dfrac{3}{\sqrt{3}}\pm2i=-\sqrt{3}\pm2i$
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