\(\displaystyle x^{2} + 4x + 4\) can be seen to fit the perfect square trinomial pattern:

\(\displaystyle x^{2} + 4x + 4 = x^{2} + 2 \cdot x \cdot 2 + 2 ^{2 } = (x + 2) ^{2 }\)

The equation can therefore be rewritten as 

\(\displaystyle \frac{8}{ (x + 2) ^{2 }} - \frac{2}{x+ 2 } - 3 = 0\)

Multiply both sides of the equation by the least common denominator of the expressions, which is \(\displaystyle LCM (x+2 , (x+2) ^{2}) = (x+2)^{2}\):

\(\displaystyle \left [\frac{8}{ (x + 2) ^{2 }} - \frac{2}{x+ 2 } - 3 \right ] (x+2)^{2} = 0 \cdot (x+2)^{2}\)

\(\displaystyle \frac{8}{ (x + 2) ^{2 }}\cdot (x+2)^{2} - \frac{2}{x+ 2 }\cdot (x+2)^{2} - 3 \cdot (x+2)^{2} = 0\)

\(\displaystyle 8 - 2 (x+2) - 3 \cdot (x+2)^{2} = 0\)

Distributing and collecting like terms, we get

\(\displaystyle 8 - 2 (x+2) - 3 \cdot (x^{2} + 4x + 4 ) = 0\)

\(\displaystyle 8 - 2 \cdot x - 2 \cdot 2 - 3 \cdot x^{2} - 3\cdot 4x - 3\cdot 4 = 0\)

\(\displaystyle 8 - 2 x - 4 - 3 x^{2} -12 x - 12= 0\)

\(\displaystyle - 3 x^{2} - 2 x -12 x + 8 - 4 - 12= 0\)

\(\displaystyle - 3 x^{2} - 14 x -8= 0\)

Multiplying both sides by \(\displaystyle -1\):

\(\displaystyle \left (- 3 x^{2} - 14 x -8 \right )(-1)= 0 (-1)\)

\(\displaystyle 3 x^{2} + 14 x +8 = 0\)

This can be solved using the \(\displaystyle ac\) method. We are looking for two integers whose sum is \(\displaystyle 14\) and whose product is \(\displaystyle 3 \times 8 = 24\). Through some trial and error, the integers are found to be 2 and 12, so the above equation can be rewritten, and solved using grouping, as

\(\displaystyle 3 x^{2} + 12x + 2x +8 = 0\)

\(\displaystyle (3 x^{2} + 12x) + (2x +8 )= 0\)

\(\displaystyle 3x (x+4 ) + 2 (x+4 )= 0\)

\(\displaystyle (3x + 2 )(x+4 )= 0\)

By the Zero Product Principle, one of these factors is equal to zero:

Either: 

\(\displaystyle 3x + 2 = 0\)

\(\displaystyle 3x + 2 - 2 = 0 - 2\)

\(\displaystyle 3x = -2\)

\(\displaystyle \frac{3x }{3}= \frac{-2}{3}\)

\(\displaystyle x = -\frac{2}{3}\)

Or:

\(\displaystyle x+ 4 = 0\)

\(\displaystyle x+ 4 - 4= 0 - 4\)

\(\displaystyle x = -4\)

Both solutions can be confirmed by substitution; the solution set is \(\displaystyle \left \{ -4, - \frac{2}{3} \right \}\)

A fourth-degree, or quartic, polynomial has four zeroes, if a zero of multiplicity \(\displaystyle n\) is counted \(\displaystyle n\) times. Since its lead term is \(\displaystyle x^{4}\), we know that

\(\displaystyle p(x) = (x-b_{1}) (x-b_{2}) (x-b_{3})(x-b_{4})\)

A polynomial with rational coefficients has its imaginary zeroes in conjugate pairs. Since \(\displaystyle p(x)\) is such a polynomial, then, since \(\displaystyle 2-3i\) is a zero, so is its complex conjugate \(\displaystyle 2+3i\); similarly, since \(\displaystyle 3+ 2 i\) is a zero, so is its complex conjugate \(\displaystyle 3- 2 i\). Substituting these four values for the four \(\displaystyle b_{n}\) values: 

\(\displaystyle p(x) = [x-(2+3i)][x-(2-3i)] [x-(3+2i)] [x-(3-2i)]\)

This can rewritten as

\(\displaystyle p(x) =[x- 2+3i][x- 2-3i] [x- 3+2i][x- 3-2i]\)

or

\(\displaystyle p(x) =[(x- 2)+3i][(x- 2)-3i] [(x- 3)+2i][(x- 3)-2i]\)

Multiply the first two factors using the difference of squares pattern, then the square of a binomial pattern:

\(\displaystyle [(x- 2)+3i][(x- 2)-3i]\)

\(\displaystyle = (x- 2)^{2}-(3i)^{2}\)

\(\displaystyle = (x- 2)^{2}-9i^{2}\)

\(\displaystyle = x^{2} -4x +4 +9\)

\(\displaystyle = x^{2} -4x +13\)

Multiply the last two factors similarly:

\(\displaystyle [(x- 3)+2i][(x- 3)-2i]\)

\(\displaystyle = (x- 3)^{2}-(2i)^{2}\)

\(\displaystyle = (x- 3)^{2}-4i^{2}\)

\(\displaystyle = x^{2} -6x+9 +4\)

\(\displaystyle = x^{2} -6x +13\)

Thus,

\(\displaystyle p(x) = ( x^{2} -4x +13) ( x^{2} -6x +13)\)

Multiply:

                            \(\displaystyle x^{2} -4x +13\)

                            \(\displaystyle \underline{ x^{2} -6x +13}\)

                       \(\displaystyle 13x^{2} -52x +169\)

       \(\displaystyle -6 x^{3} +24x^{2} -78x\)

\(\displaystyle \underline{x^{4} -4x^{3} +13x^{2}}\)________________

\(\displaystyle x^{4} - 10x^{3} + 50x^{2} - 130x +169\)

Set \(\displaystyle y = x^{\frac{1}{2}}\). Then \(\displaystyle x = \left (x^{\frac{1}{2}} \right )^{2} = y ^{2}\). 

\(\displaystyle 8 x -2 x ^{\frac{1}{2}} - 3 = 0\) can be rewritten as

\(\displaystyle 8 \left (x^{\frac{1}{2}} \right )^{2} -2 x ^{\frac{1}{2}} - 3 = 0\)

Substituting \(\displaystyle y^{2}\) for \(\displaystyle x\) and \(\displaystyle y\) for \(\displaystyle x^{\frac{1}{2}}\), the equation becomes 

\(\displaystyle 8y^{2} -2 y - 3 = 0\),

a quadratic equation in \(\displaystyle y\).

This can be solved using the \(\displaystyle ac\) method. We are looking for two integers whose sum is \(\displaystyle -2\) and whose product is \(\displaystyle 8 (-3) = -24\). Through some trial and error, the integers are found to be \(\displaystyle -6\) and 4, so the above equation can be rewritten, and solved using grouping, as

\(\displaystyle 8y^{2}-6y + 4 y - 3 = 0\)

\(\displaystyle 2y ( 4 y - 3)+1 ( 4 y - 3) = 0\)

\(\displaystyle (2y +1) ( 4 y - 3) = 0\)

By the Zero Product Principle, one of these factors is equal to zero:

Either:

\(\displaystyle 2y + 1 = 0\)

\(\displaystyle 2y + 1 - 1 = 0 - 1\)

\(\displaystyle 2y = - 1\)

\(\displaystyle \frac{2y}{2} = \frac{- 1}{2}\)

\(\displaystyle y = - \frac{1}{2}\)

Substituting back:

\(\displaystyle x^{\frac{1}{2}} = -\frac{1}{2}\), or \(\displaystyle \sqrt{x} = - \frac{1}{2}\)

However, this does not hold for any real value of \(\displaystyle x\). No solution is yielded here.

Or:

\(\displaystyle 4y - 3 = 0\)

\(\displaystyle 4y - 3+ 3 = 0+ 3\)

\(\displaystyle 4 y = 3\)

\(\displaystyle \frac{4 y}{4} = \frac{3}{4}\)

\(\displaystyle y= \frac{3}{4}\)

Substituting back:

\(\displaystyle x^{\frac{1}{2}} = \frac{3}{4}\), or \(\displaystyle \sqrt{x} = \frac{3}{4}\)

\(\displaystyle (\sqrt{x} )^{2} = \left (\frac{3}{4} \right )^{2}\)

\(\displaystyle x = \frac{9}{16}\),

the only solution. This is not among the choices.