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 $0\leq x<\frac{3 + \sqrt5}{2}$
 A: Hint: it must be $$x\geq 0$$ so we get two cases:
$$\sqrt{x}\geq x$$ or by squaring
$$x(x-1)\le 0$$ this is only possible for $$0\le x\le 1$$ and we get
$$\sqrt{x}<x<1$$ and by squaring
$$0<x^2+x+1$$ which is true.
So we get
$$0\le x\le 1$$
the second case is for you!
A: Ok, so the other $2$ solutions have minor problems with them, so here is mine:
Let $\sqrt{x}=a\geq 0$. The inequality then becomes
$$|a-a^2|<1.$$
Square both sides (by non-negativity) to get
$$(a-a^2)^2<1$$
$$(a-a^2-1)(a-a^2+1)<0$$
which has solutions when $$-a^2+a-1<0, -a^2+a+1>0$$
or when $$-a^2+a-1>0, -a^2+a+1<0.$$
The first one has solutions $a\in \left(\frac{1-\sqrt{5}}{2}, \frac{1+\sqrt{5}}{2}\right)$ while the second one has no real solutions.
Taking into account that $a\geq 0$ we have that $a \in \left[0,\frac{1+\sqrt{5}}{2}\right)$.
Square this to get $x \in \left[0,\frac{3+\sqrt{5}}{2}\right)$.
A: 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
$\frac{1}{2}-\frac{\sqrt{5}}{2} <\sqrt{x}<\frac{1}{2}+\frac{\sqrt{5}}{2}$
$0\leq x<(\frac{1}{2}+\frac{\sqrt{5}}{2})^2=\frac{3+\sqrt{5}}{2}$
A: Of course, we must immediately have that $x\ge 0$ since otherwise, we would be saying that the field of complex numbers is ordered, which it is not.
Now under that assumption we have that $\sqrt x-x<1$ or $\sqrt x-x>-1.$
The first inequality gives $\sqrt x<1+x.$ Clearly, $x+1>0$ for all $x\ge0,$ and with the convention that $\sqrt x\ge0$ we can square both sides of the inequality $\sqrt x<1+x$ and rearrange to obtain $x^2+x+1>0,$ which is clearly true for all $x\ge 0.$
The second inequality $\sqrt x-x>-1$ gives $\sqrt x>x-1.$ Now, it is clear that $x-1>0$ whenever $x>1,$ so that we can square both sides and rearrange again to obtain $x^2-3x+1<0$ which gives $$\left(x-\frac{3+\sqrt5}{2}\right)\left(x-\frac{3-\sqrt5}{2}\right)<0.$$ Therefore, it follows that either $$0\le x<\frac{3-\sqrt5}{2} \tag{*}$$ or $$x>\frac{3+\sqrt5}{2}. \tag{**}$$ On the other hand whenever $x\in[0,1],$ it is clear that $\sqrt x>x-1$ is satisfied since then $x-1<0$ (again on the convention that $\sqrt x\ge0$). Combining the inequality $0\le x\le 1$ with $(*)$ and $(**)$, we obtain $(*)$ since $(**)$ and $[0,1]$ intersect in $\varnothing.\blacksquare$
