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Solution

For the quadratic $2x^2 + 6x + a = 0$: $\alpha + \beta = -\dfrac{6}{2} = -3$, and $\alpha\beta = \dfrac{a}{2}$. Now, $\dfrac{\alpha}{\beta} + \dfrac{\beta}{\alpha} = \dfrac{\alpha^2 + \beta^2}{\alpha\beta} = \dfrac{(\alpha + \beta)^2 - 2\alpha\beta}{\alpha\beta} = \dfrac{9 - 2\cdot \frac{a}{2}}{\frac{a}{2}} = \dfrac{18 - 2a}{a}.$ Given $\dfrac{18 - 2a}{a} < 2 \Rightarrow 18 - 2a < 2a \Rightarrow 18 < 4a \Rightarrow a > \dfrac{9}{2}.$ For real roots, discriminant $36 - 8a \ge 0 \Rightarrow a \le \dfrac{9}{2}.$ Hence, no real $a$ satisfies both.