Q. One of the square root of $6 + 4\sqrt{3}$ is
A) $\sqrt{3}\left( \sqrt{3} + 1 \right)$
B) $- \sqrt{3}\left( \sqrt{3} - 1 \right)$
C) $\sqrt{3}\left( - \sqrt{3} + 1 \right)$
D) None of these
Explanation
(d)
Let $\sqrt{6 + 4\sqrt{3}} = \sqrt{x} + \sqrt{y}$
$\Rightarrow 6 + 4\sqrt{3} = x + y + 2\sqrt{xy}$
$\Rightarrow \ \ x + y = 6,\ \ \ \sqrt{xy} = 2\sqrt{3}$
Now, $(x - y)^{2} = (x + y)^{2} - 4xy$
$= 36 - 4(4 \times 3)$
$= - 12 < 0$
It is not possible
Hence, square root is not possible
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.