# Inequalities involving Absolute Values and Surds

For handling inequalities involving absolute values and surds, it is required to ensure the domains before solving inequalities. the final solutions must fit in the domains.

## Example 1

Solve for $x$, $|x| \gt \sqrt{x+2}$.

\begin{align} x+2 &\ge 0 &\color{green}{\text{domain of } \sqrt{x+2} } \\ x &\ge -2 \color{green}{\cdots (1)} \\ |x| &\gt \sqrt{x+2}^2 \\ x^2 &\gt x+2 \\ x^2 – x – 2 &\gt 0 \\ (x+1)(x-2) &\gt 0 \\ x &\lt -1 \text{ or } x \gt 2 \color{green}{\cdots (2)} \\ \require{AMSsymbols} \therefore -2 &\le x \lt -1 \text{ or } x \gt 2 &\color{green}{\text{Take common part of (1) and (2)}} \end{align}

## Example 2

Solve for $x$, $|4x-1| \gt 2\sqrt{x(1-x)}$.

\displaystyle \begin{align} x(1-x) &\ge 0 &\color{green}{\text{domain of } \sqrt{x(1-x)} } \\ x(x-1) &\le 0 \\ 0 &\le x \le 1 \color{green}{\cdots (1)} \\ |4x-1|^2 &\gt 2^2 \sqrt{x(1-x)}^2 \\ 16x^2 – 8x + 1 &\gt 4x(1-x) \\ 16x^2 – 8x + 1 &\gt 4x – 4x^2 \\ 20x^2 – 12x + 1 &\gt 0 \\ (10x-1)(2x-1) &\gt 0 \\ x &\lt \frac{1}{10} \text{ or } x \gt \frac{1}{2} \color{green}{\cdots (2)} \\ \require{AMSsymbols} \therefore 0 &\le x \lt \frac{1}{10} \text{ or } \frac{1}{2} \lt x \le 1 &\color{green}{\text{Take common part of (1) and (2)}} \end{align}