[CF25]

Does anyone know the steps to get from:

x² -2 + 1/x² > 0

To this:

(x - 1/x)² > 0

I assume it is completing the square but it could not be.

[oh_1993]

2 = 2x*(1/x)

So x^2 - 2 + 1/x^2 = x^2 - 2x*(1/x) + 1/x^2

See if you can do it from there

[CF25]

(Original post by oh_1993)
2 = 2x*(1/x)

So x^2 - 2 + 1/x^2 = x^2 - 2x*(1/x) + 1/x^2

See if you can do it from there

still can't sorry. How did you get to 2 = 2x*(1/x)

[2710]

(x - 1/x)^2

Expand this and see what you get

[CF25]

(Original post by 2710)
(x - 1/x)^2

Expand this and see what you get

The original equation:

x^2 -2 + 1/x^2

[roar558]

(Original post by CF25)
still can't sorry. How did you get to 2 = 2x*(1/x)

(1/x)*x=1
hence
2=2x*(1/x)

I'll give you a hint on how to continue:

(a+b)^2=a^2+2ab+b^2

[2710]

(Original post by CF25)
The original equation:

x^2 -2 + 1/x^2

Yes, so it is just by inspection how you get from one to the other

[TenOfThem]

It is the equivalent of knowing that

[CF25]

Thanks guys, got it.

[Xenite]

Looking at that, I would probably just do it by inspection..

However, you could also try multiplying through by x^2, and then let x^2=X, and rearrange as if it was a normal quadratic, if that's easier. You'll need to substitute back in at the end, though.


Edit: Oops, looks like you've solved it while I was posting.. never mind