If $\alpha_1, \alpha_2, \alpha_3, \alpha_4$ are roots of $x^4 +(2-\sqrt{3})x^2 +2+\sqrt{3}=0$ .... Problem : 
If $\alpha_1, \alpha_2, \alpha_3, \alpha_4$ are roots of $x^4 +(2-\sqrt{3})x^2 +2+\sqrt{3}=0$ then the value of $(1-\alpha_1)(1-\alpha_2)(1-\alpha_3)(1-\alpha_4)$ is 
(a) 2$\sqrt{3}$ 
(b)  5 
(c)  1 
(d)  4 
My approach : 
Discriminant of this problem is : 
$(2-\sqrt{3})^2- 4(2+\sqrt{3}) <0$
Therefore roots are imaginary. 
Now how to consider the roots here... please suggest.. thanks

Let give a polinom $p_4(x)=a_4 x^4+b_3x^3+c_2x^2+d_1x+e$. Use formula:
$$x_1+x_2+x_3+x_4=-\frac{b}{a}$$
$$x_1x_2+x_1x_3+x_2x_3+x_1x_4+x_2x_4+x_3x_4=\frac{c}{a}$$
$$x_1x_2x_3+x_1x_2x_4+x_1x_3x_4+x_2x_3x_4=-\frac{d}{a}$$
$$x_1x_2x_3x_4=\frac{e}{a}$$
where $x_1,x_2,x_3,x_4$ is rrots of the given polinom.
For the given example, we have: $x^4+0\cdot x^3 +(2-\sqrt 3)x^2+0\cdot x + 2+\sqrt 3=0$ $\Rightarrow$ $a=-1, b=0, c=2-\sqrt 3, d=0, e=2+\sqrt 3$
$$\alpha_1+\alpha_2+\alpha_3+\alpha_4=0$$
$$\alpha_1\alpha_2+\alpha_1\alpha_3+\alpha_2\alpha_3+\alpha_1\alpha_4+\alpha_2\alpha_4+\alpha_3\alpha_4=2-\sqrt 3$$
$$\alpha_1\alpha_2\alpha_3+\alpha_1\alpha_2\alpha_4+\alpha_1\alpha_3\alpha_4+\alpha_2\alpha_3\alpha_4=0$$
$$\alpha_1x\alpha_2\alpha_3\alpha_4=2-\sqrt 3$$
$(1-\alpha_1)(1-\alpha_2)(1-\alpha_3)(1-\alpha_4)=1-(\alpha_1+\alpha_2+\alpha_3+\alpha_4)+(\alpha_1\alpha_2+\alpha_1\alpha_3+\alpha_2\alpha_3+\alpha_1\alpha_4+\alpha_2\alpha_4+\alpha_3\alpha_4)-(\alpha_1\alpha_2\alpha_3+\alpha_1\alpha_2\alpha_4+\alpha_1\alpha_3\alpha_4+\alpha_2\alpha_3\alpha_4)+\alpha_1x\alpha_2\alpha_3\alpha_4=2-\sqrt 3+2+\sqrt 3=4$
$d)$ $4$ is answer.
