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Bulgarian Spring Tournament

Algebra Difficulty 4.8 AIME Prove it Bulgaria

Solve the inequality:
$\frac{x^2 - |x - 1| - 4}{x - 4} \geq 2x - 1.$

Solution

Note that $x \neq 4$ and $|x-1| = x-1$ for $x > 1$ , otherwise $|x-1| = 1-x$ .

If $x > 1$ we get:
$\frac{x^2 - 8x + 7}{x - 4} \le 0 \iff \frac{(x - 7)(x - 1)}{x - 4} \le 0 \iff x \in (4, 7]$

If $x \le 1$ we get:
$\frac{x^2 - 10x + 9}{x - 4} \le 0 \Leftrightarrow \frac{(x - 9)(x - 1)}{x - 4} \le 0 \Leftrightarrow x \in (-\infty, 1]$

Finally $x \in (-\infty, 1] \cup (4, 7]$ . $\square$

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Source: MathNet, licensed CC-BY-4.0. Statement reproduced verbatim; metadata (topic, difficulty) added by this project.