Given equation is x2−(a−2)x−(a+1)=0 Here, a=1,b=−(a−2) and c=−(a+1) Let the roots of equation is a and p α+β=−
b
a
=a−2 and αβ=
c
a
=−(a+1) ⇒α2+β2=(α+β)2−2αβ =(a−2)2−2{−(a+1)} ⇒α2+β2=(a−2)2+2(a+1) =a2+4−4a+2a+2 =a2−2a+1+5 ⇒α2+β2=(a−1)2+5 Clearly,α2+β2≥5 So, the minimum value of α2+β2 is 5 which it atains at a = 1.