Question

One of the square roots 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

Answer 1

Answer: (D)

(a) $[\sqrt{3}(\sqrt{3}+1){]}^2=3(3+1+2\sqrt{3})$
$=12+6\sqrt{3}$
(b) $[-\sqrt{3}(\sqrt{3}-1){]}^2=3(3+1-2\sqrt{3})$
$=12-6\sqrt{3}$
(c) $[\sqrt{3}(-\sqrt{3}+1){]}^2=3(3+1-2\sqrt{3})$
$=12-6\sqrt{3}$
Hence, none of the square values of options are equal to the given value.
Alternative
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.