The value of x satisfying (x+1)/(x-1) + (x-2)/(x+2)+(x-3)/(x+3)+(x+4)/(x-4)=4 are(a)(-5+3451/2)/10(b) (5+3401/2)/8(c)(-5+3251/2)/16(d)none

Hi

The given equation is

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

Write (x+1) as (x-1)+2. So the first term becomes

1+ 2/(x-1).

Similarly, the second term becomes 1 + 2/(x-2), and so on.

Do this to all four terms on the LHS. Now the equation becomes

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

or, 4 + 2/(x-1) + 2/(x-1) + 2/(x-1) + 2/(x-1) = 4

or, 2 ( 1/(x-1) + 1/(x-2) + 1/(x-3) + 1/(x-4)) = 0

or, 1/(x-1) + 1/(x-4) + 1/(x-3) + 1/(x-2) = 0

or,

[(x-4)+(x-1)]/[x²-5x+4] + [(x-3)+(x-2)]/[x²-5x+6] = 0

or,

(2x-5)/[x²-5x+4] = -(2x-5)/[x²-5x+6]

Therefore Either 2x-5 = 0 or 1/[x²-5x+4] = -1/[x²-5x+6]

x = 5/2 or [x²-5x+6]=-[x²-5x+4]

x=5/2 or 2x²-10x+10=0

x=5/2 or x²-5x+5=0

x=5/2 or x = (5+root(5))/2

These are the only solutions. I think there is some typing error in the options but I am sure these 3 are the only solution. I have verified it using MATLAB software also.