A complete example with the sign line

A complete exercise that brings together the Type I method and the graphical intersection of the intervals on the sign line.

Example — $\sqrt{x^2+3x-4} + \dfrac{x}{2} - 2 \le 0$

Let us isolate the root: here there is no denominator containing $x$, so we move it straight to the left-hand side. $\sqrt{x^2+3x-4} \le 2 - \tfrac{x}{2} = \tfrac{4-x}{2}.$ It is a Type I. Equivalent system: $\begin{cases} x^2+3x-4 \ge 0 & \text{(CE della radice)} \\ \tfrac{4-x}{2} \ge 0 & \text{(secondo membro } \ge 0) \\ x^2+3x-4 \le \left(\tfrac{4-x}{2}\right)^2 & \text{(elevamento)} \end{cases}$

First inequality: $x^2+3x-4=(x+4)(x-1)\ge 0 \implies x\le -4 \vee x\ge 1$.

Second inequality: $4-x\ge 0 \implies x\le 4$.

Third inequality: we expand the square $x^2+3x-4 \le \tfrac{16-8x+x^2}{4}$ and multiply by $4$ (positive, direction preserved): $4x^2+12x-16 \le 16-8x+x^2 \implies 3x^2+20x-32 \le 0.$ I solve the associated equation $3x^2+20x-32=0$: $\Delta = 400+384=784=28^2, \quad x_{1,2}=\frac{-20\pm 28}{6} \implies x_1=-8,\ x_2=\tfrac{4}{3}.$ Hence $3x^2+20x-32\le 0$ for $-8\le x\le \tfrac{4}{3}$.

Intersection of the three: $\bigl(x\le -4 \vee x\ge 1\bigr) \ \wedge\ x\le 4 \ \wedge\ -8\le x\le \tfrac{4}{3}$.

Solution: $\boxed{\,-8\le x\le -4 \ \vee\ 1\le x\le \tfrac{4}{3}\,}$.

The sign line makes the intersection visible: the three conditions are drawn one above the other and the zone common to all of them is taken.

The three conditions superimposed on the sign line; in orange the intersection, $-8\le x\le -4$ and $1\le x\le \tfrac{4}{3}$.

Links

Topics: Irrational inequalities
Concepts: Existence conditions · Irrational inequality · Sign table
Methods: Irrational inequality by cases · Sign study
Skills: Interpreting a graph · Solving inequalities