Find the quadratic equation having one root : $2\sqrt{3}-4$ .
A. ${{x}^{2}}-4\sqrt{2}x-4=0$
B. ${{x}^{2}}-8x+4=0$
C. ${{x}^{2}}+4\sqrt{2}x-4=0$
D. ${{x}^{2}}+8x+4=0$

Hint: We are asked to find the quadratic equation, with the given one root. To find this first we need to know about complex and conjugate root theorem which states that: If the root given is a surd of the form of the $\left( a-\sqrt{b} \right)$ is also a root. So, here we will take $\left( -4+2\sqrt{3} \right)$ and $\left( -4-2\sqrt{3} \right)$ as two roots and find the sum of the roots $\left( \alpha +\beta \right)$ and product of the roots $\left( \alpha \beta \right)$ to find the quadratic equation with the help of standard form of quadratic equation $a{{x}^{2}}+bx+c=0$ .

Complete step-by-step solution:
In the question, we are asked to find the quadratic equation whose one root is $\left( 2\sqrt{3}-4 \right)$. To find the quadratic equation we need two roots, If the given root is complex, then its conjugate is the other root. So, if $\left( a+\sqrt{b} \right)$ is a root, then $\left( a-\sqrt{b} \right)$ is also another root.
We know that $\left( 2\sqrt{3}-4 \right)=\left( -4+2\sqrt{3} \right)$ is an irrational root so, $\left( -2\sqrt{3}-4 \right)=\left( -4-2\sqrt{3} \right)$ will be a conjugate root.
We know that the quadratic equation in general from is given by $a{{x}^{2}}+bx+c=0$ and if the denote the two roots of the equation as $\alpha ,\beta $ then the sum and product if the roots is given by ;
$\begin{align}
  & \left( \alpha +\beta \right)=\dfrac{-b}{a} \\
 & \left( \alpha \beta \right)=\dfrac{c}{a} \\
\end{align}$
We know that $\left( \alpha +\beta \right)$ where $\alpha $ and $\beta $ are two roots we can formulate the quadratic equation as;
\[\Rightarrow {{x}^{2}}-\left( \alpha +\beta \right)x+\alpha \beta =0\]
So we have
$\Rightarrow \left( \left( 2\sqrt{3}-4 \right)+\left( -2\sqrt{3}-4 \right) \right)=\alpha +\beta $
$\begin{align}
  & \Rightarrow -4+2\sqrt{3}+\left( -4-2\sqrt{3} \right)=\alpha +\beta \\
 & \Rightarrow -4-4+2\sqrt{3}-2\sqrt{3}=\alpha +\beta \\
 & \Rightarrow -8=\alpha +\beta \\
\end{align}$
We also have;
$\begin{align}
  & \Rightarrow \left( -4+2\sqrt{3} \right)\left( -4-2\sqrt{3} \right)=\alpha \beta \\
 & \Rightarrow -\left( -4+2\sqrt{3} \right)\left( 4+2\sqrt{3} \right)=\alpha \beta \\
 & \Rightarrow \left( 4-2\sqrt{3} \right)\left( 4+2\sqrt{3} \right)=\alpha \beta \\
\end{align}$
We know that $\left( a+b \right)\left( a-b \right)={{a}^{2}}-{{b}^{2}}$ , we will apply it in the above equation.
So, $\left( 4-2\sqrt{3} \right)\left( 4-2\sqrt{3} \right)=-\left( {{\left( 4 \right)}^{2}}-{{\left( 2\sqrt{3} \right)}^{2}} \right)=\alpha \beta $
$\begin{align}
  & \Rightarrow 16-12=\alpha \beta \\
 & \Rightarrow 4=\alpha \beta \\
 & \therefore \alpha \beta =4 \\
\end{align}$
We use the formula to make quadratic with sum and product of the roots as;
$\begin{align}
  & \Rightarrow {{x}^{2}}-\left( \alpha +\beta \right)x+\alpha \beta =0 \\
 & \Rightarrow \left( 1 \right){{x}^{2}}-\left( -8 \right)x+4=0 \\
 & \therefore {{x}^{2}}+8x+4=0 \\
\end{align}$
The quadratic equation of the given root is ${{x}^{2}}+8x+4=0$.
Hence, the option (D) is the correct answer.

Note: A quadratic equation is any equation that can be rearranged in standard form as $QE=a{{x}^{2}}+bx+c=0$ where x represents an unknown, and a, b, c represent known numbers, where $a\ne 0$ . A quadratic equation always has two roots, if complex roots are included and a double root is counted for two. A quadratic equation can be factored into an equivalent equation $a{{x}^{2}}+bx+c=\left( x-r \right)\left( x-s \right)$ where r and s are the zeros or roots of quadratic equation.