Question

If the equation $(x-a)(x-b)+(x-b)(x-c)+(x-c)(x-a)=0$ has real roots that are equal in magnitude and opposite in sign, then

• A. $a+b+c=0$
• B. $a+b+c<0$
• C. $a+b+c>0$
• D. None of these

Answer 1

The correct option is A $a+b+c=0$

$(x-a)(x-b)+(x-b)(x-c)+(x-c)(x-a)=0x^2-(a+b)x+ab+x^2-(b+c)x+bc+x^2-(a+c\_x+ac=03x^2-2(a+b+c)x+(ab+ab+ac)=0ifrootsareequalinmagnitudebutoppositeinsignthencoefficientofx=0\therefore a+b+c=0$

Or

This can be done using discriminant.