1. **State the problem:** We want to find the values of $a$ such that both roots of the quadratic equation $$x^2-(2a+1)x+a^2-4a+3=0$$ are greater than 0. 2. **Recall the quadratic roots conditions:** For a quadratic equation $x^2+bx+c=0$, the roots are both positive if: - The sum of roots $-b$ is positive. - The product of roots $c$ is positive. - The discriminant $\Delta = b^2 - 4c$ is non-negative (roots are real). 3. **Identify coefficients:** Here, $b=-(2a+1)$ and $c=a^2-4a+3$. 4. **Sum of roots:** $$\text{sum} = 2a+1 > 0 \implies a > -\frac{1}{2}$$ 5. **Product of roots:** $$a^2 - 4a + 3 > 0$$ Factorize: $$(a-1)(a-3) > 0$$ This inequality holds when $a < 1$ or $a > 3$. 6. **Discriminant:** $$\Delta = (2a+1)^2 - 4(a^2 - 4a + 3)$$ Simplify: $$= 4a^2 + 4a + 1 - 4a^2 + 16a - 12 = 20a - 11$$ For real roots, $$20a - 11 \geq 0 \implies a \geq \frac{11}{20} = 0.55$$ 7. **Combine all conditions:** - $a > -\frac{1}{2}$ (sum positive) - $a < 1$ or $a > 3$ (product positive) - $a \geq 0.55$ (discriminant non-negative) From these, the intersection is: - For $a < 1$: $a \geq 0.55$ and $a < 1$ gives $0.55 \leq a < 1$ - For $a > 3$: $a > 3$ 8. **Final answer:** Both roots are positive if $$a \in [0.55, 1) \cup (3, \infty)$$