SOLUTION: <pre> 1 1 3 ——————— + ——————— = ————————— t - 1 t + 2 t² + t^-2 Please help me figure out this problem. I think it may have no s

Solve for t:

   1         1          3
——————— + ——————— = —————————
 t - 1     t + 2     t2 + t-2

My graphing calculator says it has solution t = -1.390957657, but
that may be too hard to calculate by hand.  But let's try anyway:

First we need to get rid of the negative exponent on the right

            3
        —————————
         t2 + t-2

Multiply top and bottom by t2

          3(t2)        3t2
       —————————— = —————————
       (t2 + t-2)    t4 + 1

   1         1         3t2
——————— + ——————— = —————————
 t - 1     t + 2     t4 + 1


Multiply each fraction through by the LCD = (t - 1)(t + 2)(t4 + 1)

 (t + 2)(t4 + 1) + (t - 1)(t4 + 1) = 3t2(t - 1)(t + 2)

 t5 + t + 2t4 + 2 + t5 + t - t4 - 1 = 3t2(t2 + t - 2)

                  2t5 + t4 + 2t + 1 = 3t4 + 3t3 - 6t2 

     2t5 - 2t4 - 3t3 + 6t2 + 2t + 1 = 0

The only rational solutions possible are ±1, ±1/2, but none of
these turn out to be solutions.  So it cannot be solved by any 
methods of ordinary algebra.  However it DOES have one irrational
solution, approximately what I got with a scientific calculator,
x = -1.390957657.  However, your teacher cannot expect you to get
that answer by hand, unless you use tedious iterative methods.

Edwin
AnlytcPhil@aol.com