If α  and β  are the roots of the equation x2−p(x+1)−q=0,  then the value of  α2+2α+1α2+2α+1+β2+2β+1β2+2β+q  is

The given equation  is x2−px−(p+q)=0 α  and β  are the roots of the equationSum of the roots : α+β=p,  Product of the roots : αβ=−(p+q)Now (α+1)(β+1)=αβ+(α+β)+1                                =−(p+q)+p+1                    (α+1)(β+1)   =1−q→1The given expression is α2+2α+1α2+2α+q+β2+2β+1β2+2β+q =(α+1)2(α+1)2+(q-1)+(β+1)2(β+1)2+(q-1)   =(α+1) (α+1) -(β+1)+(β+1) (β+1)-(α+1)=α-βα-β=1  from 1