Inequality with the absolute value of a trinomial (|x²-5x+6| ≥ 2x-3)

Statement

Solve the inequality $|x^2-5x+6|\ge2x-3$.

Solution

It is of the type $|A|\ge B$ with $B=2x-3$, which may be negative: the cases must be distinguished.

Case (a): $2x-3<0 \iff x<\frac32$. An absolute value is always $\ge0$, so the inequality is always true: $S_a=\left(-\infty;\frac32\right)$.

Case (b): $x\ge\frac32$. Then $x^2-5x+6\ge2x-3\ \lor\ x^2-5x+6\le-(2x-3)$:

• $x^2-7x+9\ge0 \iff x\le\dfrac{7-\sqrt{13}}{2}\ \lor\ x\ge\dfrac{7+\sqrt{13}}{2}$;
• $x^2-3x+3\le0$: $\Delta=9-12<0$, never satisfied.

Intersecting the first with $x\ge\frac32$: $S_b=\left[\frac32;\dfrac{7-\sqrt{13}}{2}\right]\cup\left[\dfrac{7+\sqrt{13}}{2};+\infty\right)$.

Union $S_a\cup S_b$: $\boxed{x\le\dfrac{7-\sqrt{13}}{2}\ \lor\ x\ge\dfrac{7+\sqrt{13}}{2}}$.

Links

Topics: Parabola
Concepts: Absolute value inequalities · Absolute value
Skills: Reasoning by cases · Solving inequalities
Exercise type: Solving an inequality