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If two eqns (ax^+bx+c=0 & a1x^2+b1x+c1=0) have a common root then, (bc1-b1c)(ab1-a1b)=(ca1-c1a)^2 here,a=a1=1 b=b,b1=c c=c,c1=b =>(b^2-c^2)(c-b)=(c-b)^2 =>b^2-c^2=c-b =>b+c=-1 =>b+c+1=0

Let A be the common root and P,Q be the uncommon roots. therefore b = - (A+P), c = AP. also c = -(A+Q) and b = AQ. now AP = - (A+Q) => A = -Q/P+1. similarly A = -P/Q+1. so P(P+1) = Q(Q+1) => P=Q or P = -(Q+1) ...... but P is not equal to Q or else b=c. this gives A = -P/(Q+1) = 1. hence P+Q = -1. hence b+c+1= -(2A+P+Q)+1 = -(2-1)+1 = -1+1 = 0.