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# Find the value of $x$ from the equation:$\sqrt 3 x - 2 = 2\sqrt 3 + 4$.(A) $x = 2\left( {1 + \sqrt 3 } \right)$ (B) $x = \left( {1 + 2\sqrt 3 } \right)$ (C) $x = 3\left( {2 + \sqrt 2 } \right)$ (D) $x = - 2\left( {3 + \sqrt 3 } \right)$  Hint: Separate the terms containing $x$ on one side and constant on the other. Evaluate the value of $x$ and rationalize the final value if required.

According to the question, the given equation is:
$\Rightarrow \sqrt 3 x - 2 = 2\sqrt 3 + 4$
Separating the terms containing variable on one side and constants on the other and then solving it further, we’ll get:
$\Rightarrow \sqrt 3 x = 2\sqrt 3 + 4 + 2, \\ \Rightarrow \sqrt 3 x = 2\sqrt 3 + 6, \\ \Rightarrow \sqrt 3 x = 2\sqrt 3 + 2 \times 3 \\$
We know that 3 can also be written as $\sqrt 3 \times \sqrt 3$, using this we’ll get:
$\Rightarrow \sqrt 3 x = 2\sqrt 3 + 2 \times \sqrt 3 \times \sqrt 3 , \\ \Rightarrow \sqrt 3 x = \sqrt 3 \left( {2 + 2\sqrt 3 } \right), \\ \Rightarrow x = 2 + 2\sqrt 3 , \\ \Rightarrow x = 2\left( {1 + \sqrt 3 } \right) \\$
So, the value of $x$ in the above equation is $2\left( {1 + \sqrt 3 } \right)$. (A) is the correct option.

Note: If in any expression, we are getting an irrational number in denominator then we can always rationalize the number to get a rational number in denominator. In rationalization, we multiply both numerator and denominator by the conjugate of denominator.
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