x2+px+2=0<rs and x2+rx+s=0<pq r+s=−p .....(1) rs=q ......(2) p+q=−r ........(3) pq=s ........(4) (2)×(4)⇒ pqrs =qs⇒pr=1 .......(5) (1)+(3)⇒p+q+r+s=−p−r ⇒2(p+r)+q+s=0 ......(6) Now p is a root of x2+rx+s=0⇒p2+rp+s=0 And r is a root of x2+px+q=0⇒r2+pr+q=0 Adding those two ⇒(p+r)2−2(p+r)=0⇒p+r=0,p+r=2 Also pr=1⇒p=1,r=1 From (4)pq=s⇒q=s ∴q+s=−2(p+r) ⇒q+s=−4⇒q=s=−2 ∴4×1−2+2−2=2