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 160: 63


The solutions are $x=\dfrac{1}{3}\pm\dfrac{\sqrt{7}}{3}$

Work Step by Step

$3x^{2}-2x-2=0$ Take out common factor $3$ from the left side: $3\Big(x^{2}-\dfrac{2}{3}x-\dfrac{2}{3}\Big)=0$ Take the $3$ to divide the right side: $x^{2}-\dfrac{2}{3}x-\dfrac{2}{3}=\dfrac{0}{3}$ $x^{2}-\dfrac{2}{3}x-\dfrac{2}{3}=0$ Take $\dfrac{2}{3}$ to the right side: $x^{2}-\dfrac{2}{3}x=\dfrac{2}{3}$ Add $\Big(\dfrac{b}{2}\Big)^{2}$ to both sides of the equation. In this case, $b=-\dfrac{2}{3}$: $x^{2}-\dfrac{2}{3}x+\Big(-\dfrac{2}{2\cdot3}\Big)^{2}=\dfrac{2}{3}+\Big(-\dfrac{2}{2\cdot3}\Big)^{2}$ $x^{2}-\dfrac{2}{3}x+\dfrac{1}{9}=\dfrac{2}{3}+\dfrac{1}{9}$ $x^{2}-\dfrac{2}{3}x+\dfrac{1}{9}=\dfrac{7}{9}$ Factor the left side of the equation, which is a perfect square trinomial: $\Big(x-\dfrac{1}{3}\Big)^{2}=\dfrac{7}{9}$ Take the square root of both sides: $\sqrt{\Big(x-\dfrac{1}{3}\Big)^{2}}=\pm\sqrt{\dfrac{7}{9}}$ $x-\dfrac{1}{3}=\pm\dfrac{\sqrt{7}}{3}$ Solve for $x$: $x=\dfrac{1}{3}\pm\dfrac{\sqrt{7}}{3}$
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