Question

If the quadratic equation \((\lambda + 2)x^2 - 3\lambda x + 4\lambda = 0, \lambda \neq -2\), has two positive roots, then the number of possible integral values of \(\lambda\) is:

• A. 1
• B. 2
• C. 3
• D. 4

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
For a quadratic equation to have two positive real roots, three conditions must hold: the discriminant must be non-negative (\(\Delta \ge 0\)), the sum of roots must be positive (\(S>0\)), and the product of roots must be positive (\(P>0\)).
Step 2: Key Formula or Approach:
For \(Ax^2 + Bx + C = 0\):
1. \(\Delta = B^2 - 4AC \ge 0\)
2. \(S = -B/A>0\)
3. \(P = C/A>0\)
Step 3: Detailed Explanation:
Given \(A = \lambda + 2\), \(B = -3\lambda\), \(C = 4\lambda\).
Condition 1: \(\Delta \ge 0\)
\[ (-3\lambda)^2 - 4(\lambda + 2)(4\lambda) \ge 0 \] \[ 9\lambda^2 - 16\lambda^2 - 32\lambda \ge 0 \implies -7\lambda^2 - 32\lambda \ge 0 \] \[ 7\lambda^2 + 32\lambda \le 0 \implies \lambda(7\lambda + 32) \le 0 \] So, \(\lambda \in \left[ -\frac{32}{7}, 0 \right]\).
Condition 2: Sum \(S>0\)
\[ \frac{3\lambda}{\lambda + 2}>0 \] Using the wavy curve method, \(\lambda \in (-\infty, -2) \cup (0, \infty)\).
Condition 3: Product \(P>0\)
\[ \frac{4\lambda}{\lambda + 2}>0 \] This yields the exact same intervals as the sum: \(\lambda \in (-\infty, -2) \cup (0, \infty)\).
Find the intersection of all conditions:
Intersection of \(\left[ -\frac{32}{7}, 0 \right]\) and \((-\infty, -2) \cup (0, \infty)\) is \(\left[ -\frac{32}{7}, -2 \right)\).
Since \(-\frac{32}{7} \approx -4.57\), the interval is approximately \([-4.57, -2)\).
The integral values in this range are \(-4\) and \(-3\).
Step 4: Final Answer:
There are 2 integral values.