If α, β , where α<β , are the roots of the equation λ x^ 2 -(λ+3) x+3=0 such that (1)/(α)-(1)/(β)=(1)/(3) , then the sum of all possible values of λ is

Correct answer: B. 6 α + β = (λ+3)/(λ) , αβ = (3)/(λ) . (1)/(α) - (1)/(β) = (β - α)/(αβ) = (1)/(3) , so β - α = (αβ)/(3) = (1)/(λ) . (β - α)^2 = (α+β)^2 - 4αβ : (1)/(λ^2) = ((λ+3)^2)/(λ^2) - (12)/(λ) 1 = (λ+3)^2 - 12λ = λ^2 - 6λ + 9 λ^2 - 6λ + 8 = 0 λ = 2 or λ = 4 . Both values satisfy α < β (verified). Sum = 2 + 4 = 6 .

JEE Main 202628 JanuaryMorning Shift

Exam: JEE Main
Year: 2026
Date: 28 January
Shift: Morning Shift
Subject: Mathematics
Chapter: Quadratic Equation

JEE Main 2026 · 28 January · Morning Shift · Mathematics · Quadratic Equation

If α, β , where α<β , are the roots of the equation λ x^ 2 -(λ+3)...

\alpha, \beta

, where

\alpha<\beta

, are the roots of the equation

\lambda x^{2}-(\lambda+3) x+3=0

such that

\frac{1}{\alpha}-\frac{1}{\beta}=\frac{1}{3}

, then the sum of all possible values of

\lambda

Options

Correct answer

B. 6

Step-by-step solution

\alpha + \beta = \frac{\lambda+3}{\lambda}

\alpha\beta = \frac{3}{\lambda}

\frac{1}{\alpha} - \frac{1}{\beta} = \frac{\beta - \alpha}{\alpha\beta} = \frac{1}{3}

, so

\beta - \alpha = \frac{\alpha\beta}{3} = \frac{1}{\lambda}

(\beta - \alpha)^2 = (\alpha+\beta)^2 - 4\alpha\beta

\frac{1}{\lambda^2} = \frac{(\lambda+3)^2}{\lambda^2} - \frac{12}{\lambda}

1 = (\lambda+3)^2 - 12\lambda = \lambda^2 - 6\lambda + 9

\lambda^2 - 6\lambda + 8 = 0 \Rightarrow \lambda = 2

\lambda = 4

.
Both values satisfy

\alpha < \beta

(verified). Sum

= 2 + 4 = 6

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