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 a quadratic equation $Ax^2 + Bx + C = 0$, the roots are given by $$x = \frac{-B \pm \sqrt{B^2 - 4AC}}{2A}.$$ Here, $A = a-1$, $B = -2$, and $C = 5$. 3. **Conditions for roots:** - Both roots must be real, so the discriminant must be non-negative: $$\Delta = B^2 - 4AC \geq 0.$$ - Both roots must be greater than 2. 4. **Calculate the discriminant:** $$\Delta = (-2)^2 - 4(a-1)(5) = 4 - 20(a-1) = 4 - 20a + 20 = 24 - 20a.$$ For real roots, $$24 - 20a \geq 0 \implies 20a \leq 24 \implies a \leq \frac{24}{20} = 1.2.$$ 5. **Express roots using the quadratic formula:** $$x = \frac{2 \pm \sqrt{24 - 20a}}{2(a-1)} = \frac{2 \pm \sqrt{24 - 20a}}{2(a-1)}.$$ 6. **Both roots greater than 2:** Let the roots be $x_1$ and $x_2$. Since $a-1$ is the coefficient of $x^2$, consider two cases: **Case 1: $a-1 > 0$ (i.e., $a > 1$)** - The parabola opens upward. - Both roots greater than 2 means the smaller root $> 2$. **Case 2: $a-1 < 0$ (i.e., $a < 1$)** - The parabola opens downward. - Both roots greater than 2 means the larger root $> 2$ and the smaller root $> 2$. 7. **Use Vieta's formulas:** Sum of roots: $$x_1 + x_2 = -\frac{B}{A} = \frac{2}{a-1}.$$ Product of roots: $$x_1 x_2 = \frac{C}{A} = \frac{5}{a-1}.$$ 8. **Inequalities for roots > 2:** Since both roots $> 2$, sum of roots $> 4$ and product of roots $> 4$ (because if both roots are greater than 2, their product is greater than $2 \times 2 = 4$). So, $$x_1 + x_2 > 4 \implies \frac{2}{a-1} > 4,$$ $$x_1 x_2 > 4 \implies \frac{5}{a-1} > 4.$$ 9. **Solve inequalities:** For sum: $$\frac{2}{a-1} > 4 \implies 2 > 4(a-1) \implies 2 > 4a - 4 \implies 6 > 4a \implies a < 1.5.$$ For product: $$\frac{5}{a-1} > 4.$$ Consider sign of denominator: - If $a-1 > 0$, then multiply both sides by $a-1$ (positive): $$5 > 4(a-1) \implies 5 > 4a - 4 \implies 9 > 4a \implies a < 2.25.$$ - If $a-1 < 0$, multiply by negative number reverses inequality: $$5 < 4(a-1) \implies 5 < 4a - 4 \implies 9 < 4a \implies a > 2.25.$$ 10. **Combine with domain from discriminant:** Recall from step 4: $a \leq 1.2$ for real roots. - For $a > 1$, sum inequality: $a < 1.5$ (true for $1 < a \leq 1.2$) - For $a > 1$, product inequality: $a < 2.25$ (true for $1 < a \leq 1.2$) - But discriminant requires $a \leq 1.2$. So for $a > 1$, both inequalities and discriminant hold if $$1 < a \leq 1.2.$$ - For $a < 1$, product inequality requires $a > 2.25$ which contradicts $a < 1$, so no solution here. 11. **Check the sign of $a-1$ in the interval $1 < a \leq 1.2$:** Here, $a-1 > 0$, parabola opens upward. 12. **Check roots greater than 2 explicitly:** Since parabola opens upward and both roots are real, both roots greater than 2 means the smaller root $> 2$. Calculate roots at $a=1.2$: $$A = 1.2 - 1 = 0.2,$$ $$\Delta = 24 - 20(1.2) = 24 - 24 = 0,$$ Root: $$x = \frac{2}{2 \times 0.2} = \frac{2}{0.4} = 5 > 2.$$ At $a=1.1$: $$A = 0.1,$$ $$\Delta = 24 - 20(1.1) = 24 - 22 = 2,$$ Roots: $$x = \frac{2 \pm \sqrt{2}}{2 \times 0.1} = \frac{2 \pm 1.414}{0.2}.$$ Smaller root: $$\frac{2 - 1.414}{0.2} = \frac{0.586}{0.2} = 2.93 > 2.$$ Thus, for $1 < a \leq 1.2$, both roots are greater than 2. **Final answer:** $$\boxed{1 < a \leq 1.2}.$$