1. **State the problem:** We need to solve the inequality $$(x^2 - x - a) \cdot \left(\sqrt{x^2 - 4x + 4} - a\right) > 0.$$ 2. **Rewrite and analyze each factor:** - The first factor is $x^2 - x - a$. - The second factor is $\sqrt{(x-2)^2} - a = |x-2| - a$ because $x^2 - 4x + 4 = (x-2)^2$. 3. **Domain considerations:** - The square root is defined for all real $x$ since $(x-2)^2 \geq 0$. 4. **Sign analysis:** The product is positive if both factors are positive or both are negative. 5. **Case 1: Both factors positive** - $x^2 - x - a > 0$ - $|x-2| - a > 0 \implies |x-2| > a$ 6. **Case 2: Both factors negative** - $x^2 - x - a < 0$ - $|x-2| - a < 0 \implies |x-2| < a$ 7. **Solve $x^2 - x - a = 0$:** Using quadratic formula: $$x = \frac{1 \pm \sqrt{1 + 4a}}{2}.$$ 8. **Intervals for $x^2 - x - a$:** - Since the parabola opens upward, $x^2 - x - a > 0$ outside the roots and $< 0$ between the roots. 9. **Analyze $|x-2| > a$ and $|x-2| < a$:** - $|x-2| > a$ means $x < 2 - a$ or $x > 2 + a$. - $|x-2| < a$ means $2 - a < x < 2 + a$. 10. **Combine intervals:** - Case 1 (both positive): $$x \in (-\infty, \frac{1 - \sqrt{1 + 4a}}{2}) \cup (\frac{1 + \sqrt{1 + 4a}}{2}, \infty)$$ and $$x < 2 - a \text{ or } x > 2 + a.$$ - Case 2 (both negative): $$x \in \left(\frac{1 - \sqrt{1 + 4a}}{2}, \frac{1 + \sqrt{1 + 4a}}{2}\right)$$ and $$2 - a < x < 2 + a.$$ 11. **Final solution:** The solution set is the union of intervals where the above conditions hold simultaneously. **Note:** The exact intervals depend on the value of $a$ and require checking overlaps. **Summary:** $$\left(\left(-\infty, \frac{1 - \sqrt{1 + 4a}}{2}\right) \cup \left(\frac{1 + \sqrt{1 + 4a}}{2}, \infty\right)\right) \cap \left(( -\infty, 2 - a) \cup (2 + a, \infty)\right)$$ $$\cup$$ $$\left(\frac{1 - \sqrt{1 + 4a}}{2}, \frac{1 + \sqrt{1 + 4a}}{2}\right) \cap (2 - a, 2 + a).$$