A nice algebra problem

The solution to the equation $ax^2 + bx+c =0$ is $x = \dfrac {-b \pm \sqrt{b^2-4ac}}{2a}$ . If there is only one solution, then $b^2 - 4ac = 0$ and $x = - \dfrac b{2a} \implies 2ax = - b$ . Now we have:

$\begin{aligned} X & = \frac 3{b-2} - \frac {3b-2}{b^2+2b+4} - \frac {b^2+16x+12}{b^2-8} - \frac 1\blue{2(ax+1)} \\ & = \frac 3{b-2} - \frac {3b-2}{b^2+2b+4} - \frac {b^2+16x+12}{b^2-8} - \frac 1\blue{-b+2} \\ & = \frac 4{b-2} - \frac {3b-2}{b^2+2b+4} - \frac {b^2+16x+12}{b^2-8} \\ & = \frac {4(b^2+2b+4) - (3b-2)(b-2)}{(b-2)(b^2+2b+4)} - \frac {b^2+16x+12}{b^2-8} \\ & = \frac {b^2+16x+12}{b^2-8} - \frac {b^2+16x+12}{b^2-8} \\ & = \boxed 0 \end{aligned}$

If there is only one solution to $ax^2 + bx + c = 0$ , then the discriminant is $0$ and $x = -\frac{b}{2a}$ , which means $-2(ax + 1) = -2(a(-\frac{b}{2a}) + 1) = b - 2$ . Then:

$\dfrac{3}{b - 2} - \dfrac{3b - 2}{b^2 + 2b + 4} - \dfrac{b^2 + 16b + 12}{b^3 - 8} - \dfrac{1}{2(ax + 1)}$

$= \dfrac{3}{b - 2} - \dfrac{3b - 2}{b^2 + 2b + 4} - \dfrac{b^2 + 16b + 12}{b^3 - 8} + \dfrac{1}{b - 2}$

$= \dfrac{4}{b - 2} - \dfrac{3b - 2}{b^2 + 2b + 4} - \dfrac{b^2 + 16b + 12}{b^3 - 8}$

$= \dfrac{4(b^2 + 2b + 4)}{(b - 2)(b^2 + 2b + 4)} - \dfrac{(3b - 2)(b - 2)}{(b - 2)(b^2 + 2b + 4)} - \dfrac{b^2 + 16b + 12}{b^3 - 8}$

$= \dfrac{4b^2 + 8b + 16}{b^3 - 8} - \dfrac{3b^2 - 8b + 4}{b^3 - 8} - \dfrac{b^2 + 16b + 12}{b^3 - 8}$

$= \dfrac{(4b^2 + 8b + 16) - (3b^2 - 8b + 4) - (b^2 + 16b + 12)}{b^3 - 8}$

$= \dfrac{4b^2 + 8b + 16 - 3b^2 + 8b - 4 - b^2 - 16b - 12}{b^3 - 8}$

$= \dfrac{4b^2 - 3b^2 - b^2 + 8b + 8b - 16b + 16 - 4 - 12}{b^3 - 8}$

$= \boxed{0}$

Since the equation has one root, $b^2=4ac$

The given expression is

$\dfrac {3}{b-2}-\dfrac {3b-2}{b^2+2b+4}-\dfrac {b^2+16b+12}{b^3-8}-\dfrac {1}{2(ax+1)}$

$=\dfrac {2ax+b}{2(b-2)(ax+1)}$

It's not mentioned that $x$ is the root of the given equation, what all is told about it is that $ax\neq -1$ . So, we have to find the maximum of the expression obtained. Since $b\neq 2$ , the only condition for the optimum of the expression is $a=0$ . Since $b^2=4ac$ , therefore $b=0$ , and the maximum of the expression is $\boxed 0$ .