Answer
$x_{1} = \dfrac{2}{9}$ and $x_{2} = -\dfrac{4}{9}$
Work Step by Step
Given $(9n+1)^2 = 9 \longrightarrow (9n)^2 + 2(9n)(1) + 1^2 = 9 \longrightarrow 81n^2 + 18n + 1 - 9 = 0 \longrightarrow 81n^2 + 18n - 8 = 0$
$a = 81, \ b = 18, \ c = -8$
Using the quadratic formula: $\dfrac{-b \pm \sqrt{b^2-4ac}}{2a} , $ we have:
$\dfrac{-18\pm \sqrt{18^2-4\times 81\times (-8)}}{2\times 81} = \dfrac{-18\pm \sqrt{324+2592}}{162} = \dfrac{-18\pm \sqrt{2916}}{162} = \dfrac{-18\pm 54}{162} = \dfrac{-1\pm 3}{9}$
Therefore, the solutions are $x_{1} = \dfrac{-1+3}{9} = \dfrac{2}{9}$ and $x_{2} = \dfrac{-1-3}{9} = -\dfrac{4}{9}$
