If $(3+2\sqrt{2})$ is a root of the equation $ax^2 + bx + c = 0$, and $(4+2\sqrt{3})$ is a root of the equation $ay^2 + my + n = 0$, where $a, b, c, m$ and $n$ are integers,
then the value of $(\frac{b}{m} + \frac{c-2b}{n})$ is

When a quadratic equation with integer coefficients has an irrational root containing a square root, the conjugate of that root must also be a root.

If we have $a + b\sqrt{k}$ as a root where $a, b$ are rational and $k$ is not a perfect square, then $a - b\sqrt{k}$ must also be a root. This ensures that when we expand the quadratic, all coefficients remain integers since the irrational parts cancel out.

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For the first equation: $ax^2 + bx + c = 0$

Given root: $(3+2\sqrt{2})$

Missing root: $(3-2\sqrt{2})$ (conjugate)

For the second equation: $ay^2 + my + n = 0$

Given root: $(4+2\sqrt{3})$

Missing root: $(4-2\sqrt{3})$ (conjugate)

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First equation with roots $(3+2\sqrt{2})$ and $(3-2\sqrt{2})$:

Sum of roots: $(3+2\sqrt{2}) + (3-2\sqrt{2}) = 6$

Product of roots: $(3+2\sqrt{2})(3-2\sqrt{2})$

Using the identity $(a+b)(a-b) = a^2 - b^2$:

$= 3^2 - (2\sqrt{2})^2 = 9 - 8 = 1$

Therefore: $x^2 - 6x + 1 = 0$

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Second equation with roots $(4+2\sqrt{3})$ and $(4-2\sqrt{3})$:

Sum of roots: $(4+2\sqrt{3}) + (4-2\sqrt{3}) = 8$

Product of roots: $(4+2\sqrt{3})(4-2\sqrt{3})$

$= 4^2 - (2\sqrt{3})^2 = 16 - 12 = 4$

Therefore: $x^2 - 8x + 4 = 0$

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From $x^2 - 6x + 1 = 0$:

$a = 1$, $b = -6$, $c = 1$

From $x^2 - 8x + 4 = 0$:

$a = 1$, $m = -8$, $n = 4$

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Substituting into $\left(\dfrac{b}{m} + \dfrac{c-2b}{n}\right)$:

$\dfrac{b}{m} = \dfrac{-6}{-8} = \dfrac{6}{8} = \dfrac{3}{4}$

$\dfrac{c-2b}{n} = \dfrac{1-2(-6)}{4} = \dfrac{1+12}{4} = \dfrac{13}{4}$

Therefore: $\left(\dfrac{b}{m} + \dfrac{c-2b}{n}\right) = \dfrac{3}{4} + \dfrac{13}{4} = \dfrac{16}{4} = 4$

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Answer: D) 4

When we work with quadratic equations having integer coefficients and irrational roots, we should always remember that irrational roots come in conjugate pairs. This property allows us to quickly form the complete quadratic equation and find all coefficients.