solve $\sqrt{x+7}<x$ for $x\in \mathbb{R}$

Question

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.

Answer 1

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 $$x+7\ge0\ \ \ \text{and}\ \ \ x+7\lt x^2\ \ \ \text{and}\ \ \ 0\lt x.$$

Answer 2

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 : $$ \sqrt{x+7}>0 \iff \begin {cases} x\ge 0\\ x+7<x^2 \end{cases} $$