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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.

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    $\begingroup$ Yes, it makes sense. It means that the solution set is empty. Actually, you should have considered $x\geq 1$, not $x>1$. $\endgroup$ – Julien Feb 24 '13 at 11:43
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    $\begingroup$ It will make sense if complex numbers are included, but if they are not then the solution set will be empty $\endgroup$ – Arjang Feb 24 '13 at 11:44
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    $\begingroup$ Empty again, since a square root is always nonnegative. $\endgroup$ – Julien Feb 24 '13 at 11:44
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    $\begingroup$ @Arjang What does $<$ mean in $\mathbb{C}$? $\endgroup$ – Julien Feb 24 '13 at 11:45
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    $\begingroup$ @Arjang Solution set empty again, exactly the same proof. $\endgroup$ – Julien Feb 24 '13 at 11:53
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Suppose that we had a real number $x$ that was a solution to the inequality.

If it were the case that $x\geq 1$, then we would have that $x^2+2<x^2-2x+1$, and hence $x<-0.5$; this is a contradiction, so our assumption that $x\geq 1$ must have been false. Therefore, if there are any solutions to the inequality, they will have to be strictly less than 1.

However, if $x<1$, then $x-1<0$; but for any real number $x$, we have that $x^2+2\geq 0$ and so $\sqrt{x^2+2}>0$; therefore, we cannot have $\sqrt{x^2+2}<x-1$, which again is a contradiction, because we assumed that $x$ did satsify the inequality.

Therefore, there cannot be any real numbers $x$ that satisfy the inequality.

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Assuming $x\in \Bbb R$, if $x<-1$, then $x-1<0$. Therefore you'll get $0\leq \sqrt{x^2+2}<x-1<0$, because a square root is always non-negative. This imples $0<0$, which can't happen. If $x\ge 1$, as the OP pointed out, $\sqrt{x^2+2}<x-1 \iff x^2+2<x^2-2x+1\iff 2x<-1\iff x<-0.5$, then $\sqrt{x^2+2}<x-1$ can't happen either.

Therefore then the set of real numbers that satisfies that inequality is $\emptyset$.

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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.

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