The quadratic equation whose one root is $2 - \sqrt{3}$ will be:

If $x = \sqrt{6 + \sqrt{6 + \sqrt{6 + \dots \text{to } \infty}}}$,then

Let $\theta$,$0 < \theta < \pi / 2$,be an angle such that the equation $x^2 + 4x \cos \theta + \cot \theta = 0$ has equal roots for $x$. Then $\theta$ in radians is

The number of all common roots of the equation $x^4-10x^3+37x^2-60x+36=0$ and the transformed equation obtained by increasing any two distinct roots of it by $1$,keeping the other two roots fixed,is

Consider the following two statements:
$I$. Any pair of consistent linear equations in two variables must have a unique solution.
$II$. There do not exist two consecutive integers,the sum of whose squares is $365$.
Then,

The roots of the equation $(x-a)(x-a-1)+(x-a-1)(x-a-2)+(x-a)(x-a-2)=0$,where $a \in R$,are always: