If #a != b != c != 0# and the equation #x^2+ax +bc=0# and #x^2+ bx+ca=0# have a common root, then find the equation formed by the 2nd root of the 2nd equation?

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Answer 1 · 2018-06-19T12:54:18

The equation is #a+b+c=0#

The equations are

#{(x^2+ax+bc=0),(x^2+bx+ca=0):}#

Let the common root be #=t#

Then,

#{(t^2+at+bc=0.........(1)),(t^2+bt+ca=0............(2)):}#

Subtracting #(2)# from #(1)#

#<=>#, #at-bt+bc-ac=0#

#<=>#, #t(a-b))=ac-bc=c(a-b)#

#<=>#, #t=c#

Let the second root of the second equation be #=beta#

Then,

#{(t+beta=c+beta=-b),(betat=betac=ca):}#

#<=>#, #{(beta=-b-c),(beta=a)}#

#<=>#, #a+b+c=0#

Also,

#a^2+ab+ca=0#

#a(a+b+c)=0#

As #a!=0#, #a+b+c=0#