<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] First of all, we need to have $x+7\ge 0$.
Then, note that $\sqrt{x+7}$ is non-negative. So, since we have
$$0\le\sqrt{x+7}\lt x,$$
we have
$$0\lt x.$$
Hence, we have
$$x+7\ge0\ \ \ \text{and}\ \ \ x+7\lt x^2\ \ \ \text{and}\ \ \ 0\lt x.$$
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