## Intermediate Algebra (12th Edition)

$x=\left\{ 1-\sqrt{5},1+\sqrt{5} \right\}$
$\bf{\text{Solution Outline:}}$ To find the solutions of the given equation, $x^2-4=2x ,$ 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} x^2-2x-4=0 .\end{array} The quadratic equation above has $a= 1 , b= -2 , c= -4 .$ 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(1)(-4)}}{2(1)} \\\\ x=\dfrac{2\pm\sqrt{4+16}}{2} \\\\ x=\dfrac{2\pm\sqrt{20}}{2} .\end{array} Simplifying the radical by writing the radicand as an expression that contains a factor that is a perfect power of the index and then extracting the root of that factor result to \begin{array}{l}\require{cancel} x=\dfrac{2\pm\sqrt{4\cdot5}}{2} \\\\ x=\dfrac{2\pm\sqrt{(2)^2\cdot5}}{2} \\\\ x=\dfrac{2\pm2\sqrt{5}}{2} .\end{array} Cancelling the common factors from all the terms results to \begin{array}{l}\require{cancel} x=\dfrac{\cancel2(1)\pm\cancel2(1)\sqrt{5}}{\cancel2(1)} \\\\ x=1\pm\sqrt{5} .\end{array} The solutions are \begin{array}{l}\require{cancel} x=1-\sqrt{5} \\\\\text{OR}\\\\ x=1+\sqrt{5} .\end{array} Hence, $x=\left\{ 1-\sqrt{5},1+\sqrt{5} \right\} .$