If the roots of the equation $ax^{2}-bx+x=0$ are of the form $\dfrac {2\alpha}{2\alpha -1}$ and $\dfrac {2\alpha}{2\alpha +1}$, then the value of $\dfrac { { \left( a+c \right) }^{ 2 } }{ ab+bc-2ac } $ is ?

Assuming the intended quadratic equation is ax^2 - bx + c = 0, the sum of its roots is b/a and the product of its roots is c/a. Adding the given roots, (2a / (2a - 1)) + (2a / (2a + 1)), gives a sum of 8a^2 / (4a^2 - 1). Multiplying the given roots yields a product of 4a^2 / (4a^2 - 1). Comparing these to the sum and product formulas, we find b = 2a and c = a, which we substitute into the requested expression. The numerator (a + c)^2 becomes (2a)^2 = 4a^2, and the denominator ab + bc - 2ac becomes 2a^2 + 2a^2 - 2a^2 = 2a^2, making the final result 4a^2 / 2a^2 = 2. The correct result is 2.

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