Answer
$9{{x}^{2}}-18x-12=0$
Work Step by Step
$1-\frac{\sqrt{21}}{3}\text{ and }1+\frac{\sqrt{21}}{3}$
It can be expressed as:
$x=1+\frac{\sqrt{21}}{3}$ and $x=1-\frac{\sqrt{21}}{3}$
$\begin{align}
& x-\left( 1+\frac{\sqrt{21}}{3} \right)=0 \\
& \left( x-1 \right)-\frac{\sqrt{21}}{3}=0
\end{align}$
Or,
$\begin{align}
& x-\left( 1-\frac{\sqrt{21}}{3} \right)=0 \\
& \left( x-1 \right)+\frac{\sqrt{21}}{3}=0 \\
\end{align}$
Apply the zero-product property
$\left( \left( x-1 \right)-\frac{\sqrt{21}}{3} \right)\left( \left( x-1 \right)+\frac{\sqrt{21}}{3} \right)=0$
$\begin{align}
& \left( \left( x-1 \right)-\frac{\sqrt{21}}{3} \right)\left( \left( x-1 \right)+\frac{\sqrt{21}}{3} \right)=0 \\
& \left( x-1 \right)\left( x-1 \right)+\frac{\sqrt{21}}{3}\left( x-1 \right)-\frac{\sqrt{21}}{3}\left( x-1 \right)-{{\left( \frac{\sqrt{21}}{3} \right)}^{2}}=0
\end{align}$
Combine like terms:
${{\left( x-1 \right)}^{2}}-{{\left( \frac{\sqrt{21}}{3} \right)}^{2}}=0$
Apply the formula, ${{\left( a+b \right)}^{2}}={{a}^{2}}+{{b}^{2}}+2ab$
$\begin{align}
& {{x}^{2}}+1-2x-\left( \frac{21}{9} \right)=0 \\
& {{x}^{2}}-2x+1-\frac{21}{9}=0 \\
& {{x}^{2}}-2x+\frac{9-21}{9}=0 \\
& {{x}^{2}}-2x-\frac{12}{9}=0
\end{align}$
Multiply the equation by $9$ to clear the fraction,
$\begin{align}
& 9\left( {{x}^{2}}-2x-\frac{12}{9} \right)=0 \\
& 9{{x}^{2}}-18x-12=0
\end{align}$
