Answer
$x=\left\{ \dfrac{-1-\sqrt{2}}{3},\dfrac{-1+\sqrt{2}}{3} \right\}$
Work Step by Step
$\bf{\text{Solution Outline:}}$
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\}
.$
