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 122: 64


The solutions are $x=\dfrac{2}{3}\pm\dfrac{\sqrt{10}}{3}$

Work Step by Step

$(3x+2)(x-1)=3x$ Evaluate the product on the left side: $3x^{2}-3x+2x-2=3x$ Take $3x$ to the left side and simplify: $3x^{2}-3x+2x-3x-2=0$ $3x^{2}-4x-2=0$ Use the quadratic formula to solve this equation. The formula is $x=\dfrac{-b\pm\sqrt{b^{2}-4ac}}{2a}$. In this case, $a=3$, $b=-4$ and $c=−2$ Substitute the known values into the formula and evaluate: $x=\dfrac{-(-4)\pm\sqrt{(-4)^{2}-4(3)(-2)}}{2(3)}=\dfrac{4\pm\sqrt{16+24}}{6}=...$ $...=\dfrac{4\pm\sqrt{40}}{6}=\dfrac{4\pm2\sqrt{10}}{6}=\dfrac{4}{6}\pm\dfrac{2\sqrt{10}}{6}=\dfrac{2}{3}\pm\dfrac{\sqrt{10}}{3}$ The solutions are $x=\dfrac{2}{3}\pm\dfrac{\sqrt{10}}{3}$
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