Question

If ${\alpha }_1,{\alpha }_2,{\alpha }_{3}and{\alpha }_4$are the roots of the equation$x^4+(2-\sqrt{3})x^2+(2+\sqrt{3})=0$, then the value of $(1-{\alpha }_1)(1-{\alpha }_2)(1-{\alpha }_3)(1-{\alpha }_4)$ is

• A. $1$
• B. $4$
• C. $2+\sqrt{3}$
• D. $5$
• E. $0$

Answer 1

The correct option is D

$5$

Find the value of$(1-{\alpha }_1)(1-{\alpha }_2)(1-{\alpha }_3)(1-{\alpha }_4)$:

Given that ${\alpha }_1,{\alpha }_2,{\alpha }_{3}and{\alpha }_4$are the roots of the equation$x^4+(2-\sqrt{3})x^2+(2+\sqrt{3})=0$

As ${\alpha }_1,{\alpha }_2,{\alpha }_{3}and{\alpha }_4$ are the roots of the equation$x^4+(2-\sqrt{3})x^2+(2+\sqrt{3})=0$

$\therefore x^4+(2-\sqrt{3})x^2+(2+\sqrt{3})=(x-{\alpha }_1)(x-{\alpha }_2)(x-{\alpha }_3)(x-{\alpha }_4)$

put$x=1$in the above equation.

we get,

$$\begin{gathered} {\left(1\right)}^4+(2-\sqrt{3}){\left(1\right)}^2+(2+\sqrt{3})=(1-{\alpha }_1)(1-{\alpha }_2)(1-{\alpha }_3)(1-{\alpha }_4) \\ \Rightarrow (1-{\alpha }_1)(1-{\alpha }_2)(1-{\alpha }_3)(1-{\alpha }_4)=1+(2-\sqrt{3})+(2+\sqrt{3}) \\ \Rightarrow (1-{\alpha }_1)(1-{\alpha }_2)(1-{\alpha }_3)(1-{\alpha }_4)=5 \end{gathered}$$

Hence, the option (D) is the correct option.