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: 44


The solutions are $x=1$ and $x=-\dfrac{5}{3}$

Work Step by Step

$3x^{2}+2x=5$ Take out common factor $3$ from the left side: $3\Big(x^{2}+\dfrac{2}{3}x\Big)=5$ Take $3$ to divide the right side: $x^{2}+\dfrac{2}{3}x=\dfrac{5}{3}$ Complete the square by adding $\Big(\dfrac{b}{2}\Big)^{2}$ to both sides of the equation and simplifying. In this case, $b=\dfrac{2}{3}$ $x^{2}+\dfrac{2}{3}x+\Big(\dfrac{2}{2\cdot3}\Big)^{2}=\dfrac{5}{3}+\Big(\dfrac{2}{2\cdot3}\Big)^{2}$ $x^{2}+\dfrac{2}{3}x+\dfrac{1}{9}=\dfrac{5}{3}+\dfrac{1}{9}$ $x^{2}+\dfrac{2}{3}x+\dfrac{1}{9}=\dfrac{16}{9}$ Factor the left side of the equation, which is a perfect square trinomial: $\Big(x+\dfrac{1}{3}\Big)^{2}=\dfrac{16}{9}$ Take the square root of both sides: $\sqrt{\Big(x+\dfrac{1}{3}\Big)^{2}}=\pm\sqrt{\dfrac{16}{9}}$ $x+\dfrac{1}{3}=\pm\dfrac{4}{3}$ Solve for $x$: $x=-\dfrac{1}{3}\pm\dfrac{4}{3}$ The two solutions are: $x=-\dfrac{1}{3}+\dfrac{4}{3}=\dfrac{3}{3}=1$ $x=-\dfrac{1}{3}-\dfrac{4}{3}=-\dfrac{5}{3}$
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