Consider the equation. (p−3)x2+2(p−3)x+2p−5=0 since, b2−4ac>0 4(p−3)2−4(p−3)(2p−5)>0 (p−3)[(p−3)−(2p−5)]>0 (p−3)(p−2)<0 Therefore, $2 Now, (β−α) is maximum at β=3 and α=2. Let f(x)=−(α+β)x2+αβx+(α−β) =−5x2+6x−1 The extreme value of f(x) is calculated as,