Question

if a,b,c belongs to real no. then prove that the roots of 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

Step-by-step explanation:

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