α+β=λλ+3, αβ=λ3.
α1−β1=αββ−α=31, so β−α=3αβ=λ1.
(β−α)2=(α+β)2−4αβ:
λ21=λ2(λ+3)2−λ12
1=(λ+3)2−12λ=λ2−6λ+9
λ2−6λ+8=0⇒λ=2 or λ=4.
Both values satisfy α<β (verified). Sum =2+4=6.
If α,β, where α<β, are the roots of the equation λx2−(λ+3)x+3=0 such that α1−β1=31, then the sum of all possible values of λ is
Held on 28 Jan 2026 · Verified 6 Jul 2026.
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