# 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?

• If you could show how you got your solution, people might be able to tell what your mistake was. – Tunococ Sep 8 '13 at 13:15
• For us to be able to tell your mistakes, we must see your "standard actions" written in detail. – Vishal Gupta Sep 8 '13 at 13:15
• Your solution for the first inequality looks good. For e.g. it is easy to check $x = \frac{1}{3}$ satisfies, but is not covered in what you say are "official results". – Macavity Sep 8 '13 at 13:28
• Your solution for the first inequality is correct, and the official results for the second is correct. – Calvin Lin Sep 8 '13 at 14:13
• For the second, you got a $2\sqrt{(2-x)(10-6x)} > 11x-7$ or so on the way, and then squared it? Note that this inequality is fulfilled if $x < \frac{7}{11}$ no matter what the left hand side is (as long as that is valid), and squaring imposes a spurious restriction on the absolute values. – Daniel Fischer Sep 8 '13 at 23:33

I premise you find this matter in every precalculus textbook.

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

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.

• Could you elaborate a bit more? – Bak1139 Sep 9 '13 at 8:49
• The official solution states otherwise ; $\frac {-5}{4}\le x \lt 1$ – Bak1139 Sep 9 '13 at 14:15
• In effect $$\left[-\frac 5{4},\frac 5{3}\right] \cap \left ]-\infty,1\right[ = \left [-\frac 5{4},1 \right[$$ really ! Check on the real line ... – Tony Piccolo Sep 9 '13 at 14:26

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.