Question

If $ {{\alpha }_{1}},{{\alpha }_{2}},{{\alpha }_{3}},{{\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

Answer: (D)

$ (1-{{\alpha }_{1}})(1-{{\alpha }_{2}})(1-{{\alpha }_{3}})(1-{{\alpha }_{4}}) $ $ =(1+{{\alpha }_{1}}{{\alpha }_{2}}-{{\alpha }_{1}}-{{\alpha }_{2}})(1-{{\alpha }_{3}})(1-{{\alpha }_{4}}) $ $ =(1+{{\alpha }_{1}}{{\alpha }_{2}}-{{\alpha }_{1}}-{{\alpha }_{2}}-{{\alpha }_{3}}-{{\alpha }_{1}}{{\alpha }_{2}}{{\alpha }_{3}} $ $ +{{\alpha }_{1}}{{\alpha }_{3}}+{{\alpha }_{2}}{{\alpha }_{3}})(1-{{\alpha }_{4}}) $ $ =1+\Sigma {{\alpha }_{1}}{{\alpha }_{2}}-\Sigma {{\alpha }_{1}}{{\alpha }_{2}}{{\alpha }_{3}}-\Sigma {{\alpha }_{1}}+{{\alpha }_{1}}{{\alpha }_{2}}{{\alpha }_{3}}{{\alpha }_{4}} $ $ =1+(2-\sqrt{3})-0-0+2+\sqrt{3} $ $ =-2\sqrt{3}+1+2+\sqrt{3}=5 $