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`
This material is intended as a summary. Use your textbook for detail explanation.
Any bug, improvement, feedback then
