Question

find the root of the following equation

1/(x-1)(x-2)+1/(x-2)(x-3)+1/(x-3)(x-4)=1/6​

Answer 1

Answer: The roots are :   -2,7

Step-by-step explanation:

1/(x-1)(x-2)  +  1/(x-2)(x-3) +  1/(x-3)(x-4)  =  1/6

LCM=(x-1)(x-2)(x-3)(x-4) for denominator

(x-3)(x-4)+ (x-1)(x-4)  + (x-1)(x-2)/(x-1)(x-2)(x-3)(x-4) = 1/6.........(1)

1. (x-3)(x-4)=x^2 -7x +12
2. (x-1)(x-4) =x^2 -5x+4
3. (x-1)(x-2)=x^2-3x+2

Adding all these we get:-3x^2-15x+18 = 3(x^2 -5x+6) =3 (x-3)(x-2) by factorisation

Substituting this in (1) gives us

3 (x-3)(x-2) / (x-1)(x-2)(x-3)(x-4) = 1/6

3 / (x-1)(x-4) = 1/6

18=(x-1)(x-4)

x^2 -5x+4=18

x^2 -5x-14=0

(x+2)(x-7)=0

x=-2 or x=7

to conclude the roots are :   -2,7