Q. If $x = \sqrt{\frac{2 + \sqrt{3}}{2 - \sqrt{3}}}$, then $x^{2}(x - 4)^{2}$ is equal to

A) 7
B) 4
C) 2
D) 1
Explanation

(d)
Given, $x = \sqrt{\frac{2 + \sqrt{3}}{2 - \sqrt{3}}} = \sqrt{\frac{\left( 2 + \sqrt{3} \right)(2 + \sqrt{3})}{\left( 2 - \sqrt{3} \right)(2 + \sqrt{3})}}$
$= 2 + \sqrt{3}$
$\therefore\ \ \ x^{2}(x - 4)^{4} = \left( 2 + \sqrt{3} \right)^{2}\left( 2 + \sqrt{3} - 4 \right)^{2}$
$= \left( \sqrt{3} + 2 \right)^{2}\left( \sqrt{3} - 2 \right)^{2}$
$= \left\lbrack \left( \sqrt{3} \right)^{2} - (2)^{2} \right\rbrack^{2}$
$= ( - 1)^{2} = 1$

Dr. Abubacker
Reviewed by Dr. Abubacker

Physics Educator & NEET/JEE Specialist

Ph.D. in Physics with 19+ years of teaching experience.