## Intermediate Algebra (12th Edition)

$r=\left\{ \dfrac{13-\sqrt{163}}{3},\dfrac{13+\sqrt{163}}{3} \right\}$
$\bf{\text{Solution Outline:}}$ To find the solutions of the given equation, $26r-2=3r^2 ,$ express first 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} -3r^2+26r-2=0 \\\\ \dfrac{-3r^2+26r-2}{-1}=\dfrac{0}{-1} \\\\ 3r^2-26r+2=0 .\end{array} The quadratic equation above has $a= 3 , b= -26 , c= 2 .$ Using $x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}$ or the Quadratic Formula, then \begin{array}{l}\require{cancel} r=\dfrac{-(-26)\pm\sqrt{(-26)^2-4(3)(2)}}{2(3)} \\\\ r=\dfrac{26\pm\sqrt{676-24}}{6} \\\\ r=\dfrac{26\pm\sqrt{652}}{6} .\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} r=\dfrac{26\pm\sqrt{4\cdot163}}{6} \\\\ r=\dfrac{26\pm\sqrt{(2)^2\cdot163}}{6} \\\\ r=\dfrac{26\pm2\sqrt{163}}{6} .\end{array} Cancelling the common factors from all the terms results to \begin{array}{l}\require{cancel} r=\dfrac{\cancel2(13)\pm\cancel2(1)\sqrt{163}}{\cancel2(3)} \\\\ r=\dfrac{13\pm\sqrt{163}}{3} .\end{array} The solutions are \begin{array}{l}\require{cancel} r=\dfrac{13-\sqrt{163}}{3} \\\\\text{OR}\\\\ r=\dfrac{13+\sqrt{163}}{3} .\end{array} Hence, $r=\left\{ \dfrac{13-\sqrt{163}}{3},\dfrac{13+\sqrt{163}}{3} \right\} .$