Let α,β be the roots of the equation
x2+(3−a)x+1−2a=0
Then, sum of roots α+β=a−3
And product of roots αβ=1−2a
We know that α2+β2=(α+β)2−2αβ
∴α2+β2=(a−3)2−2(1−2a)
=a2−2a+7
=(a−1)2+6
∴ Minimum value of α2+β2=6 at a=1.
The minimum value of the sum of the squares of the roots of x2+(3−a)x=2a−1 is
Held on 26 Jul 2022 · Verified 6 Jul 2026.
6
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8
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