<s>[INST] solve $\sqrt{x+7}
solve $\sqrt{x+7}<x$ 
I tried $\sqrt{x+7}<x\\
x+7<x^2\\
x^2-x-7>0\\
x\in \left(-\infty, \dfrac{1-\sqrt{29}}{2}\right) \cup \left( \dfrac{1+\sqrt{29}}{2},+\infty\right) $
I m not sure, if this is correct method and if the solution is correct .
I look for a simple and short way.
I have studied maths up to $12$th grade.Thanks.
 [/INST] Hint:
Note that $\sqrt{x+7}$ is a real number only if $x+7\ge 0$ and, in this case,$\sqrt{x+7}$ is a positive number so that we must have $x \ge 0$.
From all these conditions you find :
$$
\sqrt{x+7}>0 \iff 
\begin {cases}
x\ge 0\\
x+7<x^2
\end{cases}
$$
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