2. Solve the equation `6 (x^2 + 1/x^2) - 25 ( x - 1/x ) + 12 = 0`

Solution:
`6(x^2+1/x^2)-25(x-1/x)+12=0`

Let `x-1/x=m`

`=>(x-1/x)^2=m^2`

`=>x^2+1/x^2-2=m^2`

`=>x^2+1/x^2=m^2 + 2`

Substituting this values in the given equation, we get
`6(m^2+2)-25m+12=0`

`=>6m^2-25m+24=0`

`6m^2-25m+24=0`

`=>6m^2-25m+24 = 0`

`=>6m^2-16m-9m+24 = 0`

`=>2m(3m-8)-3(3m-8) = 0`

`=>(3m-8)(2m-3) = 0`

`=>(3m-8) = 0" or "(2m-3) = 0`

`=>3m = 8" or "2m = 3`

`=>m = 8/3" or "m = 3/2`

Now, `x-1/x=8/3`

`=>3x^2-3=8x`

`=>3x^2-8x-3=0`

`3x^2-8x-3=0`

`=>3x^2-8x-3 = 0`

`=>3x^2+x-9x-3 = 0`

`=>x(3x+1)-3(3x+1) = 0`

`=>(3x+1)(x-3) = 0`

`=>(3x+1) = 0" or "(x-3) = 0`

`=>3x = -1" or "x = 3`

`=>x = -1/3" or "x = 3`

Now, `x-1/x=3/2`

`=>2x^2-2=3x`

`=>2x^2-3x-2=0`

`2x^2-3x-2=0`

`=>2x^2-3x-2 = 0`

`=>2x^2+x-4x-2 = 0`

`=>x(2x+1)-2(2x+1) = 0`

`=>(2x+1)(x-2) = 0`

`=>(2x+1) = 0" or "(x-2) = 0`

`=>2x = -1" or "x = 2`

`=>x = -1/2" or "x = 2`

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