What are conditions on $x$ that make roots of $y^2 + 2xy - 2 = 0$ greater than 1 in absolute value? Original question: How do we solve $\sqrt{x^2+2}<x-1$? If $x>1$, the solution is $x<-0.5$ which does not make sense. What if $x<1$?
Update: Thank you for all answers. Actually, the original question was: "What are conditions on $x$ that make the roots of the following equation in absolute value strictly greater than one?"
$$y^2 + 2xy - 2 = 0$$
The discriminant is $D = 4x^2 +8$ and so solutions are $y_1 = -x-\sqrt{x^2+2}$ and $y_2 = -x+\sqrt{x^2+2}$. That's how I arrived to the original inequality. One condition would be $-0.5<x<0.5$. However, I discarded cases when there are complex roots. 

If you're only interested in real solutions then firstly you must have $x^2 + 2 > 0$, so $x^2 > -2$ but $x^2 > 0$ for all $x \in \mathbb{R}$ so you must have $x>0$.
Assuming $x\geq1$ and squaring both sides gives $x^2 +2 < (x-1)^2$ expanding gives $x^2 + 2 < x^2 - 2x + 1$ which can be written as $1 < -2x$ and clearly there are no solutions to this since $x\geq1$. So if a solution $x$ exists, we must have $0<x<1$ but no such solution can exist because when $0<x<1$, $\sqrt{x^2+2} > 0$ yet the other side of the inequality, $x-1<0$. So no such solution in $\mathbb{R}$ exists.
