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Question

If the roots of the equation #(x-b)(x-c)+(x-c)(x-a)+(x-a)(x-b)=0# are equal then -
i) #a+b+c=0#
ii) #a+b omega+comega^2=0 #
iii)#a-b+c=0#
iv)NONE OF THIS

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Answer 1 · 2018-08-03T08:33:47

#(iv)" NONE OF THIS"#

Here,

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

#x^2-bx-cx+bc+x^2-cx-ax+ca+x^2-ax-bx#+ab=0

#=>3x^2-2(a+b+c)x+(ab +bc+ca)=0#

Comparing with quadratic eqn. # Ax^2+Bx+C=0#

#A=3 , B=-2(a+b+c) and C=ab+bc+ca#

Given that , roots of the eqn. are equal.

So ,Discriminant #Delta=B^2-4AC=0#

#:.B^2=4AC#

#=>color(red)cancelcolor(black)4(a+b+c)^2=color(red)cancelcolor(black)4(3)(ab+bc+ca)#

#=>a^2+b^2+c^2+2ab+2bc+2ca=3ab+3bc+3ca#

#=>a^2+b^2+c^2-ab-bc-ca=0#

Multiplying each term of eqn. by #2#

#=>2a^2+2b^2+2c^2-2ab-2bc-2ca=0#

#=>a^2-2ab+b^2+b^2-2bc+c^2+c^2-2ca+a^2=0#

#=>(a-b)^2+(b-c)^2+(c-a)^2=0#

#=>a-b=0,b-c=0,c-a=0#

#=>a=b=c#

This option is not given in the question.