1. **State the problem:** We want to find the values of $a$ such that both roots of the quadratic equation $$(a-1)x^2 - 2x + 5 = 0$$ are greater than 2. 2. **Recall the quadratic formula:** For an equation $Ax^2 + Bx + C = 0$, roots are given by $$x = \frac{-B \pm \sqrt{B^2 - 4AC}}{2A}$$ 3. **Identify coefficients:** Here, $A = a-1$, $B = -2$, and $C = 5$. 4. **Condition for roots to be real:** The discriminant must be non-negative: $$\Delta = B^2 - 4AC = (-2)^2 - 4(a-1)(5) = 4 - 20(a-1) \geq 0$$ Simplify: $$4 - 20a + 20 \geq 0 \Rightarrow 24 - 20a \geq 0 \Rightarrow 20a \leq 24 \Rightarrow a \leq \frac{24}{20} = 1.2$$ 5. **Condition for roots to be greater than 2:** Both roots $x_1, x_2 > 2$. 6. **Use the fact that for quadratic $Ax^2 + Bx + C=0$ with $A \neq 0$, the parabola opens up if $A>0$ and down if $A<0$.** 7. **Evaluate the quadratic at $x=2$:** $$f(2) = (a-1)(2)^2 - 2(2) + 5 = 4(a-1) - 4 + 5 = 4a - 4 - 4 + 5 = 4a - 3$$ 8. **For both roots to be greater than 2, the parabola must be positive at $x=2$ if it opens upwards ($A>0$), so:** $$f(2) > 0 \Rightarrow 4a - 3 > 0 \Rightarrow a > \frac{3}{4} = 0.75$$ 9. **Also, since $A = a-1$, for parabola to open upwards:** $$a - 1 > 0 \Rightarrow a > 1$$ 10. **Combine conditions:** - From discriminant: $a \leq 1.2$ - From $f(2) > 0$: $a > 0.75$ - From $A > 0$: $a > 1$ So the combined interval is: $$1 < a \leq 1.2$$ 11. **Check roots at boundary $a=1$: $A=0$ so not quadratic, discard.** 12. **Final answer:** Both roots are greater than 2 if and only if $$\boxed{1 < a \leq 1.2}$$