If $\alpha_1, \alpha_2, \alpha_3, \alpha_4$ are roots of $x^4 +(2-\sqrt{3})x^2 +2+\sqrt{3}=0$ ....

Question

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

Answer 1

Hint:

$$p(x)=x^4 +(2-\sqrt{3})x^2 +2+\sqrt{3}=(x-\alpha_1)(x-\alpha_2)(x-\alpha_3)(x-\alpha_4)$$

Answer 2

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.