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

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lfryerda | High School Teacher | (Level 2) Educator

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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


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

Look at the first:


`x-1/x=0`   multiply by x




Now the second solution:


`x-1/x=3`   multiply by x

`x^2-1=3x`   rearrange

`x^2-3x-1=0`   use quadratic formula



There are four solutions to the original equation:

`x=+-1, {3+-sqrt13}/2` 

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