Question

If a,b,c belongs to r then prove that the equation 1/(x-a)+1/(x-b)+1/(x-c)=0 are always real and cannot have roots if a=b=c.

Answer 1

1/(x-a)+1/(x-b)+1/(x-c)=0
(x-b)(x-c)+(x-a)(x-c)+(x-a)(x-b)=0
x^2-bx-cx+bc+x^2-ax-cx+ac+x^2-bx-ax+ab=0
3x^2-2(a+b+c)x+(ab+bc+ac)=0
D=B^2-4AC
=4(a+b+c)^2-4*3*(ab+bc+ac)
=4(a^2+b^2+c^2+2(ab+bc+ac)-3(ab+bc+ac))
=4(a^2+b^2+c^2-(ab+bc+ac))
which is real
also the eq does not exist if a=b=c

Answer 2

Answer:

^ correct answer

Step-by-step explanation:

look above