We have, {x}^{2}+(3-\lambda )x+2=\lambda ,\text{have roots }\alpha &\beta .
Then, f(λ)=α2+β2
=(α+β)2−2αβ=(3−λ)2−2(2−λ)=λ2−6λ+9−4+2λ=λ2−4λ+5
f′(λ)=2λ−4=0 { for the minimum value of f(\lambda )}
As f′′(λ)=2>0
⇒λ=2
JEE Main 2019 — Mathematics Algebra
The value of λ such that sum of the squares of the roots of the quadratic equation, x2+(3−λ)x+2=λ has the least value is:
Held on 10 Jan 2019 · Verified 6 Jul 2026.
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