1. **State the problem:** We need to solve the inequality $$a (x^2 - x - a) \cdot (\sqrt{x^2 - 4x + 4} - a) > 0$$ for each value of the parameter $a$. 2. **Simplify the expression inside the square root:** Note that $$x^2 - 4x + 4 = (x-2)^2,$$ so $$\sqrt{x^2 - 4x + 4} = |x-2|.$$ Thus, the inequality becomes $$a (x^2 - x - a) (|x-2| - a) > 0.$$ 3. **Analyze the factors:** The inequality is a product of three factors: $a$, $(x^2 - x - a)$, and $(|x-2| - a)$. The product is positive if an even number of factors are negative or all are positive. 4. **Consider cases based on $a$:** - **Case 1: $a = 0$** The inequality becomes $0 > 0$, which is false. So no solution. - **Case 2: $a > 0$** Rewrite inequality as $$a (x^2 - x - a) (|x-2| - a) > 0,$$ Since $a > 0$, divide both sides by $a$ (positive, so inequality direction unchanged): $$ (x^2 - x - a) (|x-2| - a) > 0.$$ - **Case 3: $a < 0$** Divide both sides by $a$ (negative, so inequality direction reverses): $$ (x^2 - x - a) (|x-2| - a) < 0.$$ 5. **Analyze the quadratic $x^2 - x - a$:** The roots are given by $$x = \frac{1 \pm \sqrt{1 + 4a}}{2}.$$ - For $a \geq -\frac{1}{4}$, roots are real. - For $a < -\frac{1}{4}$, no real roots, so $x^2 - x - a$ always positive (since leading coefficient positive). 6. **Analyze $|x-2| - a$:** - For $a > 0$, $|x-2| - a > 0$ means $|x-2| > a$, i.e., $x < 2 - a$ or $x > 2 + a$. - For $a < 0$, $|x-2| - a > 0$ always true because $|x-2| \geq 0$ and $-a > 0$. 7. **Summarize solution for $a > 0$:** Solve $$(x^2 - x - a)(|x-2| - a) > 0.$$ This product is positive when both factors are positive or both negative. - Find intervals where $x^2 - x - a > 0$ and $|x-2| - a > 0$. - Find intervals where $x^2 - x - a < 0$ and $|x-2| - a < 0$. 8. **Summarize solution for $a < 0$:** Solve $$(x^2 - x - a)(|x-2| - a) < 0.$$ Since $a < 0$, $-a > 0$, so $|x-2| - a = |x-2| + |a| > 0$ always. Thus, inequality reduces to $$x^2 - x - a < 0.$$ - For $a < -\frac{1}{4}$, no real roots, so $x^2 - x - a > 0$ always, no solution. - For $-\frac{1}{4} \leq a < 0$, solution is $$\frac{1 - \sqrt{1 + 4a}}{2} < x < \frac{1 + \sqrt{1 + 4a}}{2}.$$ 9. **Final answer:** - For $a = 0$: no solution. - For $a > 0$: solve $$(x^2 - x - a)(|x-2| - a) > 0$$ using intervals defined by roots of $x^2 - x - a$ and points $2 \pm a$. - For $a < 0$: - If $a < -\frac{1}{4}$, no solution. - If $-\frac{1}{4} \leq a < 0$, solution is $$\frac{1 - \sqrt{1 + 4a}}{2} < x < \frac{1 + \sqrt{1 + 4a}}{2}.$$