Algebra 2 Common Core

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

Chapter 4 - Quadratic Functions and Equations - Chapter Review - Page 271: 60

Answer

$x=-\dfrac{3}{4}$ OR $x=1$

Work Step by Step

In the form $x^2+bx=c,$ the given equation, $ 4x^2-x-3=0 ,$ is equivalent to \begin{align*} 4x^2-x-3+3&=0+3 \\ 4x^2-x&=3 \\\\ \dfrac{4x^2-x}{4}&=\dfrac{3}{4} \\\\ \dfrac{4}{4}x^2-\dfrac{1}{4}x&=\dfrac{3}{4} \\\\ x^2-\dfrac{1}{4}x&=\dfrac{3}{4} .\end{align*} Adding $\left( \dfrac{b}{2} \right)^2$ on both sides to complete the square of the left side results to \begin{align*} x^2-\dfrac{1}{4}x+\left(-\dfrac{1/4}{2} \right)^2&=\dfrac{3}{4}+\left(-\dfrac{1/4}{2} \right)^2 \\\\ x^2-\dfrac{1}{4}x+\left(-\dfrac{1}{8} \right)^2&=\dfrac{3}{4}+\left(-\dfrac{1}{8} \right)^2 \\\\ x^2-\dfrac{1}{4}x+\dfrac{1}{64}&=\dfrac{3}{4}+\dfrac{1}{64} \\\\ \left(x-\dfrac{1}{8}\right)^2&=\dfrac{48}{64}+\dfrac{1}{64} \\\\ \left(x-\dfrac{1}{8}\right)^2&=\dfrac{49}{64} .\end{align*} Taking the square root of both sides (Square Root Property) and then solving for the variable, the equation above is equivalent to \begin{align*} x-\dfrac{1}{8}&=\pm\sqrt{\dfrac{49}{64}} \\\\ x-\dfrac{1}{8}&=\pm\dfrac{7}{8} \\\\ x&=\dfrac{1}{8}\pm\dfrac{7}{8} \end{align*} \begin{array}{lcl} x=\dfrac{1}{8}-\dfrac{7}{8} &\text{ OR }& x=\dfrac{1}{8}+\dfrac{7}{8} \\\\ x=-\dfrac{6}{8} && x=\dfrac{8}{8} \\\\ x=-\dfrac{3}{4} && x=1 .\end{array} Hence, the solutions are $x=-\dfrac{3}{4}$ OR $x=1$.
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