If a and b are the non-zero distinct roots of x2+ax+b=0, then the least value of x2+ax+b, is

Since a, bare roots of x2+ax+b=0. Therefore,a2+a2+b=0 and, b2+ab+b=0⇒ b=−2a2 and b+a+1=0⇒ −2a2+a+1=0⇒ 2a2−a−1=0⇒a=1 or, a=−1/2.Now,a=1,⇒b=−2     [∵b+a+1=0]and, a=−1/2⇒b−−1/2But, a≠b. Therefore , a=1,b=−2. ∴ Least value of x2+ax+b is −D4i.e.  −a2−4b4=−1+84=−94