<s>[INST] Final all $x$ such that $|\sqrt{x} - x| < 1$ (simple but can’t resolve) I’m a beginner so be kind with me, I tried to resolve this for 2 hours now but always give me the same wrong result. I think I’m missing something big... 
Can someone help me with this? 
I began by assuming $x\geq 0$ (root) and then I started to make 2 systems, one with the argument $(\sqrt{x} - x) \geq 0$ and the other with the argument $<0$.
Once the systems are resolved I find a different result...
The result given by the textbook is $0\leq x<\frac{3 + \sqrt5}{2}$
 [/INST] For $x\geq 0$ you have that 
$|x-\sqrt{x}|=|(\sqrt{x}-\frac{1}{2})^2-\frac{1}{4}|<1$ if and only if
$-\frac{3}{4}<(\sqrt{x}-\frac{1}{2})^2<1+\frac{1}{4}=\frac{5}{4}$ 
So
$-\frac{\sqrt{5}}{2} <\sqrt{x}-\frac{1}{2}<\frac{\sqrt{5}}{2}$
and
$\frac{1}{2}-\frac{\sqrt{5}}{2} <\sqrt{x}<\frac{1}{2}+\frac{\sqrt{5}}{2}$
$0\leq x<(\frac{1}{2}+\frac{\sqrt{5}}{2})^2=\frac{3+\sqrt{5}}{2}$
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