(2x+p+q)r=x2+px+qx+pq x2+(p+q−2r)x+pq−pr−qr=0 Let α and β be the roots. ∴α+β=−(p+q−2r) ........(i) &αβ=pq−pr−qr ........(ii) ∵α=−β (given) ∴ in eq. (1), we get ⇒−(p+q−2r)=0 ........(iii) Now, α2+β2=(α+β)2−2αβ =(−(p+q−2r))2−2(pq−pr−qr)....( from (i) and (ii)) =p2+q2+4r2+2pq−4pr−4qr−2pq+2pr+2qr =p2+q2+4r2−2pr−2qr =p2+q2+2r(2r−p−q)... (from (iii)) =p2+q2+0 =p2+q2