Solving algebraic inequality I been struggling trying to solve this inequality by hand
$$x+\sqrt{x+3} < 0$$
how should I proceed?
 A: $$x+\sqrt { x+3 } <0\Rightarrow \begin{cases} x+3\ge 0 \\ x+3<{ x }^{ 2 } \end{cases}\Rightarrow \begin{cases} x\ge -3 \\ { x }^{ 2 }-x-3>0 \end{cases}\Rightarrow \cap \begin{cases} x\ge -3 \\ x\in \left( -\infty ;\frac { 1-\sqrt { 13 }  }{ 2 }  \right) \cup \left( \frac { 1+\sqrt { 13 }  }{ 2 } ;+\infty  \right)  \end{cases}\Rightarrow \\ \Rightarrow -3\le x<\frac { 1-\sqrt { 13 }  }{ 2 } $$
A: We have
$$\sqrt{x+3}<-x$$
We can see that $x$ must be negative, since the square root is nonnegative. Therefore, assume $x<0$. We also need $x\geq-3$ for the square root to be defined. Under these assumptions, that first inequality is equivalent to
$$x+3 < x^2$$
$$\iff x^2-x-3>0$$
$$\iff x > \frac{1+\sqrt{13}}{2} \,\,\,\,\mathrm{or}\,\,\,\, x < \frac{1-\sqrt{13}}{2}$$
Therefore, 
$$x\in [-3, 0) \cap \left(-\infty, \frac{1-\sqrt{13}}{2}\right) = \left[ -3,  \frac{1-\sqrt{13}}{2}\right)$$
A: the inequality is equivalent to
$$\sqrt{x+3}<-x$$ with $$x\geq-3$$ and $$-x>0$$ thus we have
$$-3\le x<0$$
after squaring we get
 $$0<x^2-x-3$$
can you finish?
A: Let $f(x)=x+\sqrt{x+3}$.
Thus, $f$ is an increasing function.
Also, $$f\left(\frac{1-\sqrt{13}}{2}\right)=\frac{1-\sqrt{13}}{2}+\sqrt{\frac{1-\sqrt{13}}{2}+3}=$$
$$=\frac{1-\sqrt{13}}{2}+\sqrt{\frac{14-2\sqrt{13}}{4}}=\frac{1-\sqrt{13}}{2}+\frac{\sqrt{13}-1}{2}=0.$$
But the domain gives $x\geq-3$, which gives the answer $\left[-3,\frac{1-\sqrt{13}}{2}\right)$
