You should use the following substitution such that:
`x - 1/x = t`
Changing the variable yields:
`8t^2 + 2t = 121`
Moving all terms to the left side yields:
`8t^2 + 2t -121 = 0`
You need to use quadratic formula such that:
`t_(1,2) = (-2+-sqrt(4 - 4*8*(-121)))/(2*8)`
`t_(1,2) = (-2+-sqrt(4+3872))/16`
`t_(1,2) = (-2+-sqrt3876)/16`
`t_(1,2) = (-2+-2sqrt969)/16 => t_(1,2) = (-1+-sqrt969)/8`
You need to solve for x the equations `x - 1/x = t_(1,2)` such that:
`x - 1/x =(-1+sqrt969)/8 => 8x^2 - 8 = x(-1+sqrt969)`
`8x^2 -x(-1+sqrt969) - 8 =0 `
`x_(1,2) = (sqrt969 - 1+-sqrt(970-2sqrt969 + 256))/16`
`x_(1,2) = (sqrt969 - 1+-sqrt(1226-2sqrt969))/16`
`x - 1/x = (-1-sqrt969)/8 => 8x^2 - 8 = x(-1-sqrt969)`
`8x^2 -x(-1-sqrt969) - 8 = 0 `
`x_(3,4) = (-sqrt969-1+-sqrt(1226+2sqrt969))/16`
Hence, evaluating the solutions to biquadratic equation yields `x_(1,2) = (sqrt969 - 1+-sqrt(1226-2sqrt969))/16 ; x_(3,4) = (-sqrt969 - 1+-sqrt(1226+2sqrt969))/16.`
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