If the roots of the equation x2−px+q=0 differ by unity, then

Let the roots be  a and  a + 1. Then,  α+α+1=p⇒α=p−12             (i..)and , α(α+1)=q⇒α2+α=q     (ii…)From (i) and (ii), we get  p−122+p−12=q            [On eliminating a ] ⇒p2−2p+1+2p−2=4q⇒p2=4q+1.ALTER  Let the roots be αand β. Then,  |α−β|=1⇒(α−β)2=1⇒(α+β)2−4αβ=1⇒p2−4q=1