Question

If $a_1,a_2,a_3,a_4$ are the roots of equation $x^4+x^2(2-3^{\frac{1}{2}})+2+3^{\frac{1}{2}}=0$, then the value $(1-a_1)(1-a_2)(1-a_3)(1-a_4)$ is

Answer 1

We know that if the roots of equation
$p_0{x}^n+p_1{x}^{n+}+....p_n=0$
are ${\alpha }_1,{\alpha }_2....{\alpha }_n$
Product of roots $={\alpha }_1{\alpha }_2....{\alpha }_n=\frac{p_n}{p_0}$
then if we replace x by (1-x) then the roots of the equation will be
$(1-{\alpha }_1),(1-{\alpha }_2)....(1-{\alpha }_n)$
So, replacing x by (-x+1) in the given equation
$x^4+x^2(2-\sqrt{3})+2+\sqrt{3}=0$
$(1-x{)}^4+(1-x{)}^2(2-\sqrt{3})+(2+\sqrt{3})=0$
Now roots of this equation will be $(1-a_1)(1-a_2)(1-a_3)$ and $(1-a_4)$
$\therefore$ we need to find $(1-a_1)(1-a_2)(1-a_3)(1-a_4)$ which product of all roots.
$\therefore (1-a_1)(1-a_2)(1-a_3)(1-a_4)=\frac{2+\sqrt{3}+2-\sqrt{3}+1}{1}$
$=5$.