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 $&lt;0$.
Once the systems are resolved I find a different result...

The result given by the textbook is $0\leq x&lt;\frac{3 + \sqrt5}{2}$
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

$$\sqrt{x}&lt;x&lt;1$$ and by squaring
$$0&lt;x^2+x+1$$ which is true.
So we get

$$0\le x\le 1$$
the second case is for you!
We can set $\sqrt x=t$, which implies $t\ge0$. Then

$$|\sqrt x-x|&lt;1\iff (t-t^2)^2&lt;1\iff(t^2-t-1)(t^2-t+1)&lt;0.$$

The second factor has no real roots and remains positive, and the second is negative between the roots $\dfrac{1\pm\sqrt 5}2$.

Hence

$$0\le x&lt;\left(\dfrac{1+\sqrt 5}2\right)^2.$$