Algebra 2 Common Core

Published by Prentice Hall
ISBN 10: 0133186024
ISBN 13: 978-0-13318-602-4

Chapter 4 - Quadratic Functions and Equations - 4-7 The Quadratic Formula - Practice and Problem-Solving Exercises - Page 245: 19

Answer

$x=\dfrac{2-\sqrt{10}}{3} \text{ and } x=\dfrac{2+\sqrt{10}}{3} $

Work Step by Step

Using the properties of equality, the given equation, $ 3x^2=2(2x+1) ,$ is equivalent to \begin{align*} 3x^2&=2(2x)+2(1) &\text{ (use Distributive Property)} \\ 3x^2&=4x+2 \\ 3x^2-4x-2&=0 \end{align*} In the equation above, $a= 3 ,$ $b= -4 ,$ and $c= -2 .$ Using the Quadratic Formula which is given by $x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a},$ then \begin{align*}\require{cancel}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\pm\sqrt{4\cdot10}}{6} \\\\&= \dfrac{4\pm2\sqrt{10}}{6} \\\\&= \dfrac{\cancel4^2\pm\cancel2^1\sqrt{10}}{\cancel6^3} &\text{ (divide by $2$)} \\\\&= \dfrac{2\pm\sqrt{10}}{3} \end{align*} Hence, the solutions are $ x=\dfrac{2-\sqrt{10}}{3} \text{ and } x=\dfrac{2+\sqrt{10}}{3} .$
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