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Question

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

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Answer 1 · 2018-08-03T08:33:47

$(i v) NONE OF THIS$ $(iv)" NONE OF THIS"$

Here,

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

$x^{2} - b x - c x + b c + x^{2} - c x - a x + c a + x^{2} - a x - b x$ $x^2-bx-cx+bc+x^2-cx-ax+ca+x^2-ax-bx$ +ab=0

$\Rightarrow 3 x^{2} - 2 (a + b + c) x + (a b + b c + c a) = 0$ $=>3x^2-2(a+b+c)x+(ab +bc+ca)=0$

Comparing with quadratic eqn. $A x^{2} + B x + C = 0$ $Ax^2+Bx+C=0$

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

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.