quadratic equations

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17 Oct 2009 16:00:30 IST

Subject: quadratic equations

Srujana (2780)

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Let p be the comon root ,then

f(p)=g(p)=0

Rightarrow p^2+pb+ca=0

p^2+pc+ab=0

Solving both,

p(b-c)=a(b-c)

Hence 'a' is the common root

Rightarrow a^2+ab+ca=0

Rightarrow a=0 or a+b+c=0

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