If $\alpha_{1},\ \alpha_{2},\ \alpha_{3},\ \alpha_{4}$ are the roots of the equation $x^{4} + \left( 2 - \sqrt{3} \right)x^{2} + 2 + \sqrt{3} = 0$, then the value of $\left( 1 - \alpha_{1} \right)\left( 1 - \alpha_{2} \right)\left( 1 - \alpha_{3} \right)(1 - \alpha_{4})$ is

Question

Dr. Abubacker · Accepted Answer

(d)
Here, $\sum_{}^{}{\alpha_{1} = 0,\ \ \ \ \ \sum_{}^{}{\alpha_{1}\alpha_{2} = (2 - \sqrt{3})}}$,
$\sum_{}^{}{\alpha_{1}\alpha_{2}\alpha_{3} = 0,\ \ \ \sum_{}^{}{\alpha_{1}\alpha_{2}\alpha_{3}\alpha_{4} = 2 + \sqrt{3}}}$
$\left( 1 - \alpha_{1} \right)\left( 1 - \alpha_{2} \right)\left( 1 - \alpha_{3} \right)(1 - \alpha_{4})$
$= \left( 1 + \alpha_{1}\alpha_{2} - \alpha_{1} - \alpha_{2} \right)\left( 1 - \alpha_{3} \right)(1 - \alpha_{4})$
$= \left( 1 + \alpha_{1}\alpha_{2} - \alpha_{1} - \alpha_{2} - \alpha_{3} - \alpha_{1}\alpha_{2}\alpha_{3} + \alpha_{1}\alpha_{3} + \alpha_{2}\alpha_{3} \right)(1 - \alpha_{4})$
$= 1 + \sum_{}^{}{\alpha_{1}\alpha_{2} - \sum_{}^{}{\alpha_{1}\alpha_{2}\alpha_{3} - \sum_{}^{}{\alpha_{1} + \alpha_{1}\alpha_{2}\alpha_{3}\alpha_{4}}}}$
$= 1 + 2 - \sqrt{3} - 0 - 0 + 2 + \sqrt{3} = 5$

Q. If $\alpha_{1},\ \alpha_{2},\ \alpha_{3},\ \alpha_{4}$ are the roots of the equation $x^{4} + \left( 2 - \sqrt{3} \right)x^{2} + 2 + \sqrt{3} = 0$, then the value of $\left( 1 - \alpha_{1} \right)\left( 1 - \alpha_{2} \right)\left( 1 - \alpha_{3} \right)(1 - \alpha_{4})$ is

Explanation

Reviewed by Dr. Abubacker

Want more practice?