Answer with Explanation choose correct option ?

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

1 Answer
Aug 3, 2018

(iv)" NONE OF THIS"

Explanation:

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.