Irrational inequalities question: $\sqrt { -3x+1 } + \sqrt {6x+1} < \sqrt {3x+4}$ and $\sqrt { -6x+10 } + \sqrt {-x+2} \gt \sqrt {4x+5}$ Consider the following inequalities: 
\begin{equation*}
 \sqrt { -3x+1 } + \sqrt {6x+1} \lt \sqrt  {3x+4}, \\
 \sqrt { -6x+10 } + \sqrt {-x+2} \gt \sqrt {4x+5}.
\end{equation*}
Attempt at a solution; after performing all the standard actions, I got: $\frac 16\lt x \le\frac 13 $  or  $\frac {-1}{6}\le x \lt 0 $, for first inequality, and:  $\frac {-34}{97}\lt x \lt 1 $  for the second. 
Official results state however:  $\frac {-1}{6}\le x \lt 0 $  for first inequality, and
 $\frac {-5}{4}\le x \lt 1 $ for the second. 
What was my mistake? 
 A: I premise you find this matter in every precalculus textbook.
About the second inequality:
The point is that the set of the solutions of $$\sqrt A>B$$ is given by the union of the sets of the solutions of the two systems $$\begin {cases} A \ge 0 \\ \\ B<0 \end {cases}\quad ,\quad \begin {cases} B \ge 0 \\ \\  A>B^2 \end {cases}$$
Use this after the first squaring.
Of course at the start you must find the set of the admissible solutions, solving $$\begin {cases} -6x+10 \ge 0 \\ \\ -x+2 \ge 0 \\ \\  4x+5 \ge 0 \end {cases}$$
There is no problem with the first squaring because the two sides, where they exist, are not negative.
About the first inequality: $$\sqrt A<B \quad \rightarrow \quad \begin {cases} A \ge 0 \\ \\ B>0 \\ \\ A<B^2\end {cases}$$
addendum (for the second inequality):
The set of the admissible solutions is $$\left[-\frac 5{4},\frac 5{3}\right]$$
The sets of the solutions of the two systems are $$\left]-\infty,\frac 7{11}\right[ \quad,\quad \left[-\frac 7{11},1\right[$$
Therefore their union is $$\left]-\infty,1\right[$$
The intersection of the latter and the set of the admissible solutions is the solution set.
A: The second equation became $ 97x^2-66x-31<0$
$(97x+31)(x-1)<0$
Therefore
$$-\frac{31}{97}<x<1$$
This condition satisfies all conditions.
