Example — root larger than a line

Let us see the two-case method at work on a Type II inequality.

Example — $\sqrt{x+1}\ge -6x+3$

Case 1: $-6x+3<0 \iff x>\tfrac12$. Together with $x+1\ge 0$, that is $x\ge -1$, we obtain $x>\tfrac12$.

Case 2: $\begin{cases} x+1\ge 0 \\ -6x+3\ge 0 \\ x+1 \ge (-6x+3)^2 = 36x^2-36x+9 \end{cases} \iff \begin{cases} x\ge -1 \\ x\le \tfrac12 \\ 36x^2-37x+8\le 0 \end{cases}$ The third inequality: $\Delta = 37^2-4\cdot 36\cdot 8 = 1369-1152 = 217$. The zeros are $x_{1,2} = \frac{37\pm\sqrt{217}}{72} \approx \frac{37\pm 14{,}73}{72} \implies x_1\approx 0{,}31,\quad x_2\approx 0{,}72.$ Therefore $36x^2-37x+8\le 0$ for $0{,}31 \le x \le 0{,}72$. Intersecting with $x\ge -1$ and $x\le \tfrac12 = 0{,}5$ we obtain $0{,}31 \le x \le 0{,}5$.

Union Case 1 $\cup$ Case 2: $x>\tfrac12 \ \cup\ [0{,}31,\ 0{,}5]$, which combined become $\boxed{\,x\ge 0{,}31\,}$ (formally $x\ge \frac{37-\sqrt{217}}{72}$).

The blue curve, $y=\sqrt{x+1}$, starts from $(-1;0)$ and grows slowly. The red line $y=-6x+3$ is decreasing: on the left it is above the root, on the right it ends up below. The inequality asks for the points where the blue curve is above or on the red one: the solution is the whole region starting from the intersection point $x\approx 0{,}31$.

$y=\sqrt{x+1}$ against $y=-6x+3$: the solution is the region where the root is above or on the line, starting from $x\approx 0{,}31$.

Links

Topics: Irrational inequalities
Concepts: Existence conditions · Irrational inequality · Squaring
Methods: Irrational inequality by cases
Skills: Interpreting a graph · Reasoning by cases · Solving inequalities