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: Apply Vieta's formulas.
For 2x² + 6x + k = 0, the sum of roots is α + β = -6/2 = -3, and the product is αβ = k/2.

Step 2: Rewrite the target expression.
α/β + β/α = (α² + β²)/(αβ) = [(α + β)² - 2αβ]/(αβ).

Step 3: Substitute the known values.
Plugging in: [(-3)² - 2(k/2)]/(k/2) = (9 - k)/(k/2) = 2(9 - k)/k = 18/k - 2.

Step 4: Analyze the behavior for k<0.
Since k is negative, 18/k is also negative, making 18/k - 2<-2. As k → -∞, 18/k → 0 from the negative side, so the expression approaches -2 from below. The maximum limiting value is therefore -2.

Step 5: Final conclusion.
The maximum value is -2.