1. **State the problem:** We need to solve the inequality $$(x^2 - x - a) \cdot \left(\sqrt{x^2 - 4x + 4} - a\right) > 0$$ for each value of the parameter $a$. 2. **Rewrite and analyze the expression:** Note that $$\sqrt{x^2 - 4x + 4} = \sqrt{(x-2)^2} = |x-2|.$$ So the inequality becomes: $$ (x^2 - x - a) \cdot (|x-2| - a) > 0. $$ 3. **Consider the domain:** The expression is defined for all real $x$ because the square root is always non-negative. 4. **Analyze the sign of each factor:** 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. **Analyze $x^2 - x - a$:** The quadratic $x^2 - x - a = 0$ has roots: $$x = \frac{1 \pm \sqrt{1 + 4a}}{2}.$$ - For $a \geq -\frac{1}{4}$, roots are real. - For $a < -\frac{1}{4}$, no real roots and $x^2 - x - a > 0$ for all $x$. 8. **Analyze $|x-2| > a$ or $|x-2| < a$:** - If $a < 0$, $|x-2| > a$ is always true and $|x-2| < a$ is never true. - If $a \geq 0$, $|x-2| > a$ means $x < 2 - a$ or $x > 2 + a$. - $|x-2| < a$ means $2 - a < x < 2 + a$. 9. **Summarize solution by cases:** **Case A: $a < -\frac{1}{4}$** - $x^2 - x - a > 0$ for all $x$. - If $a < 0$, $|x-2| > a$ always true. - So product $> 0$ for all $x$. **Case B: $a \geq -\frac{1}{4}$ and $a < 0$** - Roots real. - $|x-2| > a$ always true. - Product sign depends on $x^2 - x - a$. - So solution is $x < \frac{1 - \sqrt{1 + 4a}}{2}$ or $x > \frac{1 + \sqrt{1 + 4a}}{2}$. **Case C: $a \geq 0$** - Product positive if: - $x^2 - x - a > 0$ and $|x-2| > a$ (i.e., $x < 2 - a$ or $x > 2 + a$), or - $x^2 - x - a < 0$ and $|x-2| < a$ (i.e., $2 - a < x < 2 + a$). 10. **Final solution:** $$\boxed{\text{For each } a:\quad (x^2 - x - a)(|x-2| - a) > 0 \text{ holds where:} }$$ - If $a < -\frac{1}{4}$, solution is all real $x$. - If $-\frac{1}{4} \leq a < 0$, solution is $$x < \frac{1 - \sqrt{1 + 4a}}{2} \text{ or } x > \frac{1 + \sqrt{1 + 4a}}{2}.$$ - If $a \geq 0$, solution is $$\left\{x: x^2 - x - a > 0 \text{ and } (x < 2 - a \text{ or } x > 2 + a)\right\} \cup \left\{x: x^2 - x - a < 0 \text{ and } 2 - a < x < 2 + a\right\}.$$