If α and β are the real roots of the equation x2-(k-2)x+k2+3k+5=0(k∈R) Find the maximum and minimum values of α2+β2 .

Given that the equation x2−k-2x+k2+3k+5=0 has real rootsit means its discriminant must be positive k+22−4k2+3k+5≥0−3k2−16k−16≥03k2+16k+16≤03k+4k+4≤0k∈−4,−43Consider α2+β2=α+β2-2αβ =k-22-2k2+3k+5 =-k2-10k-6   Differentiate both sides and equate to zero2k=-10 k=-5but this value is not in the range of k values hence the minimum or maximum values of      α    are at k=-4,-43Therefore, the extremum values are 18,509