## Intermediate Algebra (12th Edition)

$x=\left\{ \dfrac{-1-\sqrt{2}}{3},\dfrac{-1+\sqrt{2}}{3} \right\}$
$\bf{\text{Solution Outline:}}$ To find the solutions of the given equation, $9x^2+6x=1 ,$ 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} 9x^2+6x-1=0 .\end{array} The quadratic equation above has $a= 9 , b= 6 , c= -1 .$ Using $x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}$ or the Quadratic Formula, then \begin{array}{l}\require{cancel} x=\dfrac{-6\pm\sqrt{6^2-4(9)(-1)}}{2(9)} \\\\ x=\dfrac{-6\pm\sqrt{36+36}}{18} \\\\ x=\dfrac{-6\pm\sqrt{72}}{18} \\\\ x=\dfrac{-6\pm\sqrt{36\cdot2}}{18} \\\\ x=\dfrac{-6\pm\sqrt{(6)^2\cdot2}}{18} \\\\ x=\dfrac{-6\pm6\sqrt{2}}{18} \\\\ x=\dfrac{6(-1\pm\sqrt{2})}{18} \\\\ x=\dfrac{\cancel6(-1\pm\sqrt{2})}{\cancel6(3)} \\\\ x=\dfrac{-1\pm\sqrt{2}}{3} .\end{array} The solutions are \begin{array}{l}\require{cancel} x=\dfrac{-1\pm\sqrt{2}}{3} \\\\\text{OR}\\\\ x=\dfrac{-1\pm\sqrt{2}}{3} .\end{array} Hence, $x=\left\{ \dfrac{-1-\sqrt{2}}{3},\dfrac{-1+\sqrt{2}}{3} \right\} .$