The quadratic equation whose one root is $2-\sqrt{3}$ will be
A. $x^{2}-4x-1=0$
B. $x^{2}-4x+1=0$
C. $x^{2}+4x-1=0$
D. $x^{2}+4x+1=0$

Hint:To determine the quadratic equation, first consider the other root to be a conjugate of the first one. Then find the sum of roots and the product of roots by using their respective formula. After that put their value in the formula used which includes both sum of roots and the product of roots. That will give us our required quadratic equation.

Formula Used:
Quadratic Equation: $x^{2}-\left ( \alpha +\beta \right )x+\alpha \beta $

Complete step by step Solution:
We refer to equations with a second-degree variable as quadratic equations. They are also referred to as "Equations of degree 2" for this reason. The solution or root of a quadratic equation is the value of a variable for which the equation is fulfilled.
Let us consider the roots of the required quadratic equation to be $\alpha$ and $\beta$.
One of the root’s values is already given which is,
$\alpha =2-\sqrt{3}$
Then the other root will be
$\beta =2+\sqrt{3}$
Now, we need to determine the sum and the product of roots.
That is,
$\alpha +\beta =\left ( 2-\sqrt{3} \right )+\left ( 2+\sqrt{3} \right )$
$\Rightarrow \alpha +\beta =4$
Also,
$\alpha \beta =\left ( 2-\sqrt{3} \right )\left ( 2+\sqrt{3} \right )$
$\alpha \beta =4-3=1$
Now, we can get the required quadratic equation by the given formula,
$x^{2}-\left ( \alpha +\beta \right )x+\alpha \beta $
Thus, substituting the value of and in the above equation, we get
$x^{2}-4x+1=0$

Hence, the correct option is (B).


Note: Students need to be careful while assuming the value of another root. They need to take conjugate of the given root, which should be equal in magnitude but only opposite in sign.