Precalculus (6th Edition)

Published by Pearson
ISBN 10: 013421742X
ISBN 13: 978-0-13421-742-0

Chapter 1 - Equations and Inequalities - 1.4 Quadratic Equations - 1.4 Exercises - Page 121: 46


The solutions are $x=1\pm\dfrac{2\sqrt{6}}{3}$

Work Step by Step

$-3x^{2}+6x+5=0$ Divide the whole equation by $-3$: $-\dfrac{1}{3}(-3x^{2}+6x+5=0)$ $x^{2}-2x-\dfrac{5}{3}=0$ Take $\dfrac{5}{3}$ to the right side: $x^{2}-2x=\dfrac{5}{3}$ Complete the square by adding $\Big(\dfrac{b}{2}\Big)^{2}$ to both sides of the equation and simplifying. In this case, $b=-2$ $x^{2}-2x+\Big(-\dfrac{2}{2}\Big)^{2}=\dfrac{5}{3}+\Big(-\dfrac{2}{2}\Big)^{2}$ $x^{2}-2x+1=\dfrac{5}{3}+1$ $x^{2}-2x+1=\dfrac{8}{3}$ Factor the left side of the equation, which is a perfect square trinomial: $(x-1)^{2}=\dfrac{8}{3}$ Take the square root of both sides: $\sqrt{(x-1)^{2}}=\pm\sqrt{\dfrac{8}{3}}$ $x-1=\pm\dfrac{\sqrt{24}}{3}$ $x-1=\pm\dfrac{2\sqrt{6}}{3}$ Solve for $x$: $x=1\pm\dfrac{2\sqrt{6}}{3}$
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