Example — root smaller than a line

Let us apply the three-condition method to a concrete case.

Example — $\sqrt{x^2-3x+2}\le x-1$

Equivalent system: $\begin{cases} x^2-3x+2\ge 0 \\ x-1\ge 0 \\ x^2-3x+2 \le (x-1)^2 \end{cases}$

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

Second inequality: $x\ge 1$.

Third inequality: $x^2-3x+2\le x^2-2x+1 \implies -3x+2\le -2x+1 \implies -x\le -1 \implies x\ge 1.$

Intersection: $\bigl(x\le 1 \vee x\ge 2\bigr) \wedge x\ge 1 \wedge x\ge 1$. The region $x\le 1$ intersected with $x\ge 1$ gives only the point $x=1$; the region $x\ge 2$ intersected with $x\ge 1$ gives $x\ge 2$.

Solution: $\boxed{\,x=1 \ \vee\ x\ge 2\,}$.

The graph clarifies the situation: the blue curve is $y=\sqrt{x^2-3x+2}$ (it exists only for $x\le 1$ or $x\ge 2$, where it “meets” the $x$-axis); the red line is $y=x-1$. The inequality asks where the blue curve lies below or on the red one: at $x=1$ there is a pointwise contact ($0\le 0$), and then from $x=2$ onwards the line stays above the root.

The curve $y=\sqrt{x^2-3x+2}$ and the line $y=x-1$: the solution is where the root lies below or on the line, that is $x=1$ and $x\ge 2$.

Remark — Why squaring alone is NOT enough

If we limited ourselves to squaring, we would obtain $-x\le -1$, that is $x\ge 1$: a wrong solution, because it includes the whole interval $1<x<2$, where however the root does not exist (the argument is negative). The system with the three conditions serves precisely to prevent this error.

Links

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