If \[\sqrt {9{x^2}\; + {\text{ }}6x{\text{ }} + {\text{ }}1{\text{ }}} {\text{ < }}\left( {2 - x} \right)\] , then
(A) \[x \in \left( { - \dfrac{3}{2},\dfrac{1}{4}} \right)\]
(B) \[x \in \left( { - \dfrac{3}{2},\dfrac{1}{4}} \right]\]
(C) \[x \in \left[ { - \dfrac{3}{2},\dfrac{1}{4}} \right)\]
(D) \[x < \dfrac{1}{4}\]

Hint: Since the inequality given is “<” , x lies in an open interval.
In the question given, on squaring the equation we can get rid of the square root. Upon factorizing and solving for x we can determine the values of x.

Complete step by step Solution:
We are given that,
\[\sqrt {9{x^2}\; + {\text{ }}6x{\text{ }} + {\text{ }}1{\text{ }}} {\text{ < }}\left( {2 - x} \right)\]
Squaring on both sides we get,
\[9{x^2}\; + {\text{ }}6x{\text{ }} + {\text{ }}1{\text{ < }}{\left( {2 - x} \right)^2}\]
Expanding the R.H.S we get,
\[ \Rightarrow \,\,9{x^2}\; + {\text{ }}6x{\text{ }} + {\text{ }}1{\text{ < 4 - 4x + }}{{\text{x}}^2}\]
\[\; \Rightarrow 8{x^2} + 10x - 3 < 0\]
On splitting the middle term to factorize the above equation we get,
\[\;8{x^2} + 12x - 2x - 3 < 0\]
\[ \Rightarrow \;4x(2x + 3) - 1(2x + 3) < 0\]
\[ \Rightarrow (4x - 1)(2x + 3) < 0\]
\[ \Rightarrow - \dfrac{3}{2} < x < \dfrac{1}{4}\]
\[ \Rightarrow x \in \left( { - \dfrac{3}{2},\dfrac{1}{4}} \right)\]

Hence, the correct option is A.

Note:In order to solve the given question, one must know to expand and factorize equations.
The symbol \[a < x < b\] means “is an element of”. Therefore \[x \in A\] indicates that x is an element of the set A. We write that \[x \in (a,b)\] if \[a < x < b\].