## Intermediate Algebra (12th Edition)

$x=\left\{ \dfrac{1-\sqrt{3}}{2},\dfrac{1+\sqrt{3}}{2} \right\}$
$\bf{\text{Solution Outline:}}$ To find the solutions of the given equation, $2x^2-2x=1 ,$ express in the form $ax^2+bx+c=0.$ Then use the Quadratic Formula. $\bf{\text{Solution Details:}}$ Using the properties of equality, in the form $ax^2+bx+c=0,$ the expression above is equivalent to \begin{array}{l}\require{cancel} 2x^2-2x-1=0 .\end{array} The quadratic equation above has $a= 2 , b= -2 , c= -1 .$ Using $x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}$ or the Quadratic Formula, then \begin{array}{l}\require{cancel} x=\dfrac{-(-2)\pm\sqrt{(-2)^2-4(2)(-1)}}{2(2)} \\\\ x=\dfrac{2\pm\sqrt{4+8}}{4} \\\\ x=\dfrac{2\pm\sqrt{12}}{4} \\\\ x=\dfrac{2\pm\sqrt{4\cdot3}}{4} \\\\ x=\dfrac{2\pm\sqrt{(2)^2\cdot3}}{4} \\\\ x=\dfrac{2\pm2\sqrt{3}}{4} \\\\ x=\dfrac{2(1\pm\sqrt{3})}{4} \\\\ x=\dfrac{\cancel2(1\pm\sqrt{3})}{\cancel2(2)} \\\\ x=\dfrac{1\pm\sqrt{3}}{2} .\end{array} The solutions are \begin{array}{l}\require{cancel} x=\dfrac{1-\sqrt{3}}{2} \\\\\text{OR}\\\\ x=\dfrac{1+\sqrt{3}}{2} .\end{array} Hence, $x=\left\{ \dfrac{1-\sqrt{3}}{2},\dfrac{1+\sqrt{3}}{2} \right\} .$