Question

If \(\alpha,\beta\) are the roots of the equation \[ 2x^2+6x+k=0, \] then the maximum value of \[ \frac{\alpha}{\beta}+\frac{\beta}{\alpha} \] when \(k<0\) is:

• A. \(0\)
• B. \(1\)
• C. \(-1\)
• D. \(-2\)

Hint:

For expressions like \(\frac{\alpha}{\beta}+\frac{\beta}{\alpha}\), convert them into \(\frac{\alpha^2+\beta^2}{\alpha\beta}\), then use sum and product of roots.

Answer 1

The Correct Option is D

Step 1: Use sum and product of roots.
For the quadratic equation \[ 2x^2+6x+k=0, \] we have \[ \alpha+\beta=-\frac{6}{2}=-3 \] and \[ \alpha\beta=\frac{k}{2}. \]

Step 2: Express the given expression in terms of \(\alpha+\beta\) and \(\alpha\beta\).
We need \[ \frac{\alpha}{\beta}+\frac{\beta}{\alpha}. \] This can be written as \[ \frac{\alpha^2+\beta^2}{\alpha\beta}. \] Now, \[ \alpha^2+\beta^2=(\alpha+\beta)^2-2\alpha\beta. \] Therefore, \[ \frac{\alpha}{\beta}+\frac{\beta}{\alpha} = \frac{(\alpha+\beta)^2-2\alpha\beta}{\alpha\beta}. \]

Step 3: Substitute the values.
Using \[ \alpha+\beta=-3 \] and \[ \alpha\beta=\frac{k}{2}, \] we get \[ \frac{\alpha}{\beta}+\frac{\beta}{\alpha} = \frac{(-3)^2-2\left(\frac{k}{2}\right)}{\frac{k}{2}}. \] \[ = \frac{9-k}{\frac{k}{2}}. \] \[ = \frac{2(9-k)}{k}. \] \[ = \frac{18}{k}-2. \]

Step 4: Use the condition \(k<0\).
Since \[ k<0, \] we have \[ \frac{18}{k}<0. \] Therefore, \[ \frac{18}{k}-2<-2. \] As \(k\) becomes very large negative, \(\frac{18}{k}\) approaches \(0\) from the negative side.
Hence, the maximum limiting value is \[ -2. \]

Step 5: Final conclusion.
Therefore, \[ \boxed{-2} \]