Find the value of $(1-a_1)(1-a_2)(1-a_3)(1-a_4)$ given that $a_1,a_2,a_3,a_4$ are roots of a biquadratic equation. So the biquadratic equation is $x^4+(2-\sqrt3)x^2+2+\sqrt3=0$. Let $a_1,a_2,a_3,a_4$ be its roots. So we have to find the value of $(1-a_1)(1-a_2)(1-a_3)(1-a_4)$ . 
My attempt: 
So of we put $x^2=t$, and let the roots of the new quadratic equation be $a_1,a_2$. So we get that $a_1=-a_3;a_2=-a_4$. So the expression whose value we have to find will become $(1-a_1^2)(1-a_2^2)$. And now we multiply the expression to get, $1-(a_1^2+a_2^2)+(a_1a_2)^2=1-((a_1+a_2)^2-2a_1a_2)+(a_1a_2)^2$. 
Now substituting the values of sum of roots and product of the same, I get the answer $5+10\sqrt3$. But the answer according to my question paper is $5$. 
So please can someone point out the mistake and tell the answer?
Hint:

Just notice that
$$(x-a_1)(x-a_2)(x-a_3)(x-a_4)=x^4+(2-\sqrt{3})x^2+2+\sqrt{3}$$



About your attempt:

According to G-man (thanks) we have

\begin{align}
1-(a_1^2+a_2^2)+(a_1a_2)^2&amp;=1-\left(-\frac{2-\sqrt{3}}{1}\right)+(2+\sqrt{3})\\
&amp;=1+(2-\sqrt{3})+(2+\sqrt{3})\\
&amp;=5
\end{align}
Where we have used the fact that $(a_1a_2)^2=a_1a_2a_3a_4=2+\sqrt{3}$ from the  Vieta's formulas.
\begin{align}
f(x) &amp;= (x-a_1)(x-a_2)(x-a_3)(x-a_4) \\
&amp;= x^4 +(2−√3) x^2 +2 + √3 \\
f(1)&amp; = 1 + 2 - √3 + 2 - √3 = 5
\end{align}