# `(x^2+1/x^2)-3(x-1/x)=0` can some one solve this (lesson quadratic equations)!!! Thank You

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To solve the equation `x^2+1/x^2-3(x-1/x)-2=0` we can use a very interesting trick that comes up occasionally. Notice that the first two terms can be combined with the last term to get

`x^2-2+1/x^2-3(x-1/x)=0` now let `u=x-1/x`

but this means that `u^2=x^2-2+1/x^2` since the x's cancel out in the middle term.

The equation now becomes

`u^2-3u=0` factor

`u(u-3)=0`

which has two solutions u=0 and u=3.

Look at the first:

`u=0`

`x-1/x=0` multiply by x

`x^2-1=0`

`x^2=1`

`x=+-1`

Now the second solution:

`u=3`

`x-1/x=3` multiply by x

`x^2-1=3x` rearrange

`x^2-3x-1=0` use quadratic formula

`x={3+-sqrt{9+4}}/2`

`={3+-sqrt13}/2`

**There are four solutions to the original equation:**

**`x=+-1, {3+-sqrt13}/2` **