Intermediate Algebra (6th Edition)

Published by Pearson
ISBN 10: 0321785045
ISBN 13: 978-0-32178-504-6

Chapter 8 - Section 8.3 - Solving Equations by Using Quadratic Methods - Exercise Set - Page 502: 52


$-3-3i\sqrt3, -3+3i\sqrt3, \text{ and }6$

Work Step by Step

Use the Zero Factor Property by equating each factor to zero, then solve each equation: \begin{array}{ccc} &x-6=0 &\text{or} &x^2+6x+36=0 \\&x=6 & \text{or} &x^2+6x+36=0 \end{array} Solve the second equation using the quadratic formula with $a=1, b=6, c=36$ to obtain: $x=\dfrac{-6\pm \sqrt{6^2-4(1)(36)}}{2(1)} \\x=\dfrac{-6\pm \sqrt{36-144}}{2} \\x=\dfrac{-6\pm \sqrt{-108}}{2} \\x=\dfrac{-6 \pm \sqrt{36(-3)}}{2} \\x=\dfrac{-6 \pm 6\sqrt{-3}}{2} \\x=\dfrac{-6\pm 6i\sqrt{3}}{2} \\x=\dfrac{-6}{2}\pm \dfrac{6i\sqrt{3}}{2} \\x=-3\pm3i\sqrt{3}$ Therefore, the solutions of the equation are: $-3-3i\sqrt3, -3+3i\sqrt3, \text{ and }6$
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