# Solve for x. `8 (x - 1/x)^2 + 2(x + 1/x) = 121`this is a question from chapter quadratic eqations

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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.` **

t cat be x - 1/x there are two different terms x-1/x and x+1/x