# Chapter 1 - Section 1.4 - Quadratic Equations - 1.4 Exercises - Page 112: 63

$x=\left\{ \dfrac{-3-\sqrt{41}}{8},\dfrac{-3+\sqrt{41}}{8} \right\}$

#### Work Step by Step

$\bf{\text{Solution Outline:}}$ To solve the given equation, $(4x-1)(x+2)=4x ,$ use the FOIL Method and express the equation in the form $ax^2+bx+c=0.$ Then use the Quadratic Formula. $\bf{\text{Solution Details:}}$ Using the FOIL Method which is given by $(a+b)(c+d)=ac+ad+bc+bd,$ the expression above is equivalent to\begin{array}{l}\require{cancel} 4x(x)+4x(2)-1(x)-1(2)=4x \\\\= 4x^2+8x-x-2=4x \\\\= 4x^2+(8x-x-4x)-2=0 \\\\= 4x^2+3x-2=0 .\end{array} In the equation above, $a= 4 ,$ $b= 3 ,$ and $c= -2 .$ Using the Quadratic Formula which is given by $x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a},$ then \begin{array}{l}\require{cancel} x=\dfrac{-3\pm\sqrt{3^2-4(4)(-2)}}{2(4)} \\\\ x=\dfrac{-3\pm\sqrt{9+32}}{8} \\\\ x=\dfrac{-3\pm\sqrt{41}}{8} .\end{array} The solutions are $x=\left\{ \dfrac{-3-\sqrt{41}}{8},\dfrac{-3+\sqrt{41}}{8} \right\} .$

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