As a, b, c, p, q $$\in$$ R and the two given equations have exactly one common root.

$$\Rightarrow$$ Either both equations have real roots.

Or both equations have imaginary roots.

$$\Rightarrow$$ Either $$\Delta_1\ge0$$ and $$\Delta_2 > 0$$ or

$$\Delta_1 < 0$$ and $$\Delta_2 < 0$$

$$\Rightarrow~p^2-q\ge0$$ and $$b^2-ac\ge0$$

or $$p^2-q\le0$$ and $$b^2-ac\le0$$

$$\Rightarrow~(p^2-q)(b^2-ac)\ge0$$

$$\therefore$$ Statement 1 is true.

Also, we have $$\alpha\beta=q$$ and $${\alpha \over \beta } = {c \over a}$$

$$\therefore$$ $${{\alpha \beta } \over {{\alpha \over \beta }}} = {q \over c} \times a \Rightarrow {\beta ^2} = {{qa} \over c}$$

As $$\beta \ne 1$$ or $$-1$$

$$ \Rightarrow {\beta ^2} \ne 1$$

$$ \Rightarrow {{qa} \over c} \ne 1$$ or $$c \ne qa$$

Again as exactly one root $$\alpha$$ is common and $$\beta\ne1$$

$$\therefore$$ $$\alpha + \beta \ne \alpha + {1 \over \beta } \Rightarrow {{ - 2b} \over a} \ne - 2P$$

$$ \Rightarrow b \ne ap$$

$$\therefore$$ Statement 2 is correct.

But Statement 2 is not correct explanation of Statement 1.