The quadratic equations $$x^2 - 6x + a = 0$$ and $$x^2 - cx + 6 = 0$$ have one root in common. The other roots of the first and second equations are integers in the ratio 4 : 3. Then the common root is

Let the common root of the two quadratics be $$\alpha$$.

Denote the other (distinct) root of the first quadratic $$x^2 - 6x + a = 0$$ by $$p$$.
Denote the other root of the second quadratic $$x^2 - cx + 6 = 0$$ by $$q$$.

Given that the two “other” roots are integers in the ratio $$4:3$$, write them as
$$p = 4t, \qquad q = 3t$$
for some non-zero integer $$t$$.

For a quadratic $$x^2 - (\,\text{sum of roots}\,)x + (\,\text{product of roots}\,) = 0$$, we have

1. For the first quadratic
Sum of its roots $$\Longrightarrow \alpha + p = 6 \;$$ $$\Rightarrow$$ $$\alpha + 4t = 6$$ $$\;-(1)$$
    Product of its roots $$\Longrightarrow \alpha p = a$$ (not needed further).

2. For the second quadratic
    Sum of its roots $$\Longrightarrow \alpha + q = c$$ (value of $$c$$ not needed).
    Product of its roots $$\Longrightarrow \alpha q = 6$$ $$\Rightarrow$$ $$\alpha \cdot 3t = 6$$ $$\;-(2)$$

Use $$(2)$$ first:

From $$(2)$$, $$\alpha = \dfrac{6}{3t} = \dfrac{2}{t}$$ $$\;-(3)$$

Substitute $$(3)$$ in $$(1)$$:

$$\dfrac{2}{t} + 4t = 6$$

Multiply throughout by $$t$$:

$$2 + 4t^2 = 6t$$

Bring all terms to one side:

$$4t^2 - 6t + 2 = 0$$

Divide by $$2$$ to simplify:

$$2t^2 - 3t + 1 = 0$$

Solve the quadratic in $$t$$:

Discriminant $$D = (-3)^2 - 4(2)(1) = 9 - 8 = 1$$

Hence $$t = \dfrac{3 \pm 1}{2 \cdot 2} = \dfrac{3 \pm 1}{4}$$
$$\Longrightarrow t = 1 \; \text{or} \; t = \dfrac{1}{2}$$

But $$t$$ must be an integer (because $$p$$ and $$q$$ are integers).
Therefore $$t = 1$$.

Put $$t = 1$$ in $$(3)$$:

$$\alpha = \dfrac{2}{1} = 2$$.

Thus the common root of the two quadratic equations is $$2$$.

Option D which is: 2