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$x^2=2$

and $x=+\sqrt2$ and $x=-\sqrt2$.

That's ok.

But when $x^2>2$, (It's my fault. $x^2<2$ (inequality fault) => $x^2>2$ (ok))

$x > \pm\sqrt{2}$ ?

The answer is $x > \sqrt{2}$, and $x<-\sqrt{2}$.

When I was young, maybe I studied $x$ bigger than positive number, and smaller than negative number.

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    The notation x > \pm\sqrt{2} really doesn't make any sense. What exactly is it supposed to mean? Do you consider that statement to be true for $x=0$, for example?2012-02-13

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From $x^2\gt 2$ we can conclude that $\sqrt{x^2}\gt \sqrt{2}.$ However, the important point to remember is that $\sqrt{x^2}$ is not equal to $x$, it is equal to $|x|$, the absolute value of $x$. That is, we have $|x|\gt \sqrt{2}.$

And, by the definition of the absolute value, $|x|\gt\sqrt{2}$ if and only if $x\gt 0$ and $x\gt \sqrt{2}$, or $x\lt 0$ and $-x\gt\sqrt{2}$, which is equivalent to $x\lt-\sqrt{2}$; so $|x|\gt\sqrt{2}\text{ is equivalent to }x\gt\sqrt{2}\text{ or }x\lt-\sqrt{2}.$ You either write it as two inequalities, with an "or" connective, or as the single inequality using the absolute value.

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    aha, absolute value.. Thank you so much!! I want to be like you2012-02-13
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I think you have a typo: $x^2<2$ implies $-\sqrt2 which is fine, but the inequality $x^2>2$ means that either $x>\sqrt2$ or $x<-\sqrt2$, not both simultaneously (which is impossible). See for yourself that the graph of $y=x^2$ is above that of $y=2$ only outside the interval $[-\sqrt2,\sqrt2]$:

$\hskip 1.8in$ picture

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    @anon:sorry anon. $f(x)=x^2$ like your graph. To f(x)>2, x>\sqrt{2}, x<-\sqrt{2}. How to know it in equation? I don't know the step...2012-02-13
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If $x > \sqrt{2}$, then x² > 2. You seem to be having trouble with the difference between "smaller" and "more negative". A number's relative "size" is its absolute value, or its distance from zero.

Since the problem asks for x² < 2, the "less than" sign is unaffected and your desired $x$ is $x < \pm\sqrt{2} $.

If x<-\sqrt{2} \Rightarrow x² < (-\sqrt{2})² =2. If x < +\sqrt{2} \Rightarrow x² < (+\sqrt{2}) = 2.

Good luck.

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This is how I always viewed this. Not sure if totally correct.

$x^2 > 2$ is similar to $|x|^2 > 2$, so $|x| > \sqrt{2}$, so $x < -\sqrt{2}$ or $x > \sqrt{2}$