If x^2 + bx + c = 0 , x^2 +cx + b = 0 (b is not equal to c ) have a common root , then b + c + 1 = ?
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Answered by
11
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
(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
Answered by
2
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.
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.
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