Solve $x+3<\sqrt{x+33}$ : Solution Verification First of all the part inside the square root should be greater than or equal to $0$. So, 
$x+33 \ge0$
$\Rightarrow x \ge -33$
Now I solved the inequality,
$x+3<\sqrt{x+33}$
$\Rightarrow(x+3)^2<{x+33}$
$\Rightarrow x^2+9+6x<{x+33}$
$\Rightarrow x^2+5x-24 < 0$
$\Rightarrow (x+8)(x-3) < 0$
$\Rightarrow -8<x<3$
Now combining the two results i.e. $ x \ge -33$ and $ -8<x<3$ , the answer should be $ -8<x<3$ as per my understanding but this is not the correct answer that has been provided. What am I doing wrong? Please help me on this !!!
Thanks in advance !!!
 A: HINT
You can only square both sides iff both sides and the argument of the square root are nonnegative.
Under such conditions, the proposed inequation is equivalent to
\begin{align*}
x + 3 < \sqrt{x + 33} & \Longleftrightarrow
\begin{cases}
(x+3)^{2} < x + 33\\\\
x+3 \geq 0\\\\
x + 33 \geq 0
\end{cases}\\\\
& \Longleftrightarrow
\begin{cases}
x^{2} + 6x + 9 < x + 33\\\\
x\geq -3
\end{cases}\\\\
& \Longleftrightarrow
\begin{cases}
x^{2} + 6x - 34 < 0\\\\
x \geq -3
\end{cases}
\end{align*}
Moreover, notice whenever $-33 \leq x \leq -3$, the LHS is negative and the RHS is positive.
Hence the interval $[-33,-3]$ is also a solution.
Can you take it from here?
A: The problem is when you square. Notice that $x+3 \leq 0$ whenever $x \leq -3$. Also, $\sqrt{x+33} \geq 0$ for all values of $x \geq -33$. You should take this into consideration when evaluating the inequality.
A: For the inequality $\sqrt A> B$ we need of course $A\ge 0$, and you did it correct, but before squaring both sides we need to consider two cases:

*

*when $B<0$ the inequality is always satisfied

*when $B\ge 0$ we can square both sides to find other solutions

and you went wrong in that second step.
Therefore in general we have:

$$\sqrt A> B \iff \begin{cases}A\ge 0\\B<0 \end{cases}\quad \lor \quad \begin{cases}A\ge 0\\B>0\\A>B^2 \end{cases}$$

