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Maths Question 5 – JEE-MAIN 2026

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If the quadratic equation (λ+2)x2−3λx+4λ=0, λ≠−2, has two positive roots, then the number of possible integral values of λ is:

For a quadratic equation ax2+bx+c=0 to have two positive roots, three conditions must be satisfied: the discriminant must be non-negative, the sum of the roots must be positive, and the product of the roots must be positive.

Step 1: Identify Conditions for Positive Roots✦ Active

For the quadratic equation (λ+2)x2−3λx+4λ=0 to have two positive roots, the following conditions must hold:

1. Discriminant D≥0 (real roots).

2. Sum of roots (α+β)>0.

3. Product of roots (αβ)>0.

4. The leading coefficient (λ+2)≠0, which is given as λ≠−2.

Step 2: Apply Discriminant Condition○ Expand

The discriminant is D=(−3λ)2−4(λ+2)(4λ). Setting D≥0:

9λ2−16λ(λ+2)≥0 ⟹9λ2−16λ2−32λ≥0 ⟹−7λ2−32λ≥0 ⟹7λ2+32λ≤0 ⟹λ(7λ+32)≤0

This inequality holds for λ∈[−327,0].

💡 Teacher's Secret Hint

Remember to reverse the inequality sign when dividing by a negative number.

Step 3: Apply Sum and Product of Roots Conditions and Find Intersection○ Expand

Sum of roots: α+β=−−3λλ+2=3λλ+2>0. This implies λ∈(−∞,−2)∪(0,∞).

Product of roots: αβ=4λλ+2>0. This also implies λ∈(−∞,−2)∪(0,∞).

The intersection of all conditions is:

λ∈[−327,0]∩((−∞,−2)∪(0,∞))

Since −327≈−4.57, the interval from the discriminant is [−4.57,0]. The intersection with (−∞,−2) yields [−32/7,−2). The intersection with (0,∞) yields an empty set. Thus, the possible values for λ are in the interval [−327,−2).

The integral values of λ in this interval are −4 and −3. Therefore, the number of possible integral values of λ is 2.

💡 Teacher's Secret Hint

Carefully analyze the intersection of intervals, especially when dealing with open and closed endpoints.

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