<p>Find the values of the parameter <em>a</em> such that the roots \(\alpha\) and \(\beta\) of the equation \(2x^2 + 6x + a = 0\) satisfy the inequality \(\dfrac{\alpha}{\beta} + \dfrac{\beta}{\alpha} < 2\).</p>
Step-by-Step Solution
Key Concept: Transform the inequality $\frac{\alpha}{\beta} + \frac{\beta}{\alpha} < 2$ into $\frac{\alpha^2 + \beta^2}{\alpha\beta} < 2$, then use Vieta's formulas to express in terms of $a$. The critical constraint is that roots must be real and have the same sign for the ratio inequality to make sense.
<p><strong>Step 1:</strong> Express $\frac{\alpha}{\beta} + \frac{\beta}{\alpha}$ using Vieta's formulas.</p><p>$$\frac{\alpha}{\beta} + \frac{\beta}{\alpha} = \frac{\alpha^2 + \beta^2}{\alpha\beta} = \frac{(\alpha + \beta)^2 - 2\alpha\beta}{\alpha\beta}$$</p><p><strong>Step 2:</strong> Apply Vieta's formulas for $2x^2 + 6x + a = 0$:</p><p>$\alpha + \beta = -3$ and $\alpha\beta = \frac{a}{2}$</p><p><strong>Step 3:</strong> Substitute into the inequality:</p><p>$$\frac{(-3)^2 - 2 \cdot \frac{a}{2}}{\frac{a}{2}} < 2$$</p><p>$$\frac{9 - a}{\frac{a}{2}} < 2$$</p><p>$$\frac{2(9-a)}{a} < 2$$</p><p><strong>Step 4:</strong> Solve the inequality. For $a > 0$: $2(9-a) < 2a \Rightarrow 18 - 2a < 2a \Rightarrow 18 < 4a \Rightarrow a > \frac{9}{2}$</p><p>For $a < 0$: Inequality reverses: $2(9-a) > 2a \Rightarrow a < \frac{9}{2}$ (automatically satisfied)</p><p><strong>Step 5:</strong> Check discriminant: $\Delta = 36 - 8a \geq 0 \Rightarrow a \leq \frac{9}{2}$ (for real roots)</p><p><strong>Step 6:</strong> Combine conditions:</p><p>From Step 4: $a < 0$ or $a > \frac{9}{2}$</p><p>From Step 5: $a \leq \frac{9}{2}$</p><p>∴ Answer: $a < 0$ or $a > \frac{9}{2}$</p>
Correct Answer: a < 0 or a > 9/2