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Quadratic with Two Positive Roots: 2 Integer λ Values | JEE 2026

JEE Maths question with a full step-by-step solution.

Question
If the quadratic equation (λ+2)x2−3λx+4λ=0(\lambda+2)x^2-3\lambda x+4\lambda=0, λ≠−2\lambda\ne-2, has two positive roots, then the number of possible integral values of λ\lambda is
A11
B22correct
C33
D44
Solution
Step 1 (C-1: same-sign product means a⋅f(0)>0a\cdot f(0)>0): (λ+2)(4λ)>0⇒λ<−2(\lambda+2)(4\lambda)>0\Rightarrow\lambda<-2 or λ>0\lambda>0. Step 2 (C-2: vertex to the right of 0, −b2a>0-\tfrac{b}{2a}>0): 3λ2(λ+2)>0⇒λ<−2\dfrac{3\lambda}{2(\lambda+2)}>0\Rightarrow\lambda<-2 or λ>0\lambda>0. Step 3 (C-3: real roots, D≥0D\ge0): (−3λ)2−4(λ+2)(4λ)≥0⇒λ(7λ+32)≤0⇒λ∈[−327,0].(-3\lambda)^2-4(\lambda+2)(4\lambda)\ge0\Rightarrow \lambda(7\lambda+32)\le0\Rightarrow\lambda\in\left[-\dfrac{32}{7},0\right]. Step 4: Intersection of C-1, C-2, C-3:
λ∈[−327,−2)≈[−4.57, −2).\lambda\in\left[-\frac{32}{7},-2\right)\approx[-4.57,\,-2).
 Step 5: Integer values of λ\lambda in this interval are λ=−4\lambda=-4 and λ=−3\lambda=-3, i.e. 22 values. Correct answer: (2)
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