1. **State the problem:** We want to find for which values of $a$ the roots of the quadratic equation $$(a+2)x^2 - 4x + 1 = 0$$ are less than 3. 2. **Use the function and evaluate at $x=3$:** Define $$f(x) = (a+2)x^2 - 4x + 1.$$ We evaluate $$f(3) = (a+2) \cdot 3^2 - 4 \cdot 3 + 1 = 9(a+2) - 12 + 1 = 9a + 18 - 11 = 9a + 7.$$ 3. **Interpretation:** If both roots are less than 3, then $x=3$ is to the right of both roots. Since the parabola opens upwards if $a+2 > 0$ and downwards if $a+2 < 0$, the sign of $f(3)$ helps determine the position of roots relative to 3. 4. **Check the sign of $a+2$:** - If $a+2 > 0$, parabola opens upwards. - If $a+2 < 0$, parabola opens downwards. 5. **Condition for roots less than 3:** - For $a+2 > 0$, $f(3) > 0$ means the parabola is above the x-axis at 3, so roots are less than 3. - For $a+2 < 0$, $f(3) < 0$ means the parabola is below the x-axis at 3, so roots are less than 3. 6. **Solve inequalities:** - For $a+2 > 0$, $9a + 7 > 0 \Rightarrow 9a > -7 \Rightarrow a > -\frac{7}{9}$. - For $a+2 < 0$, $9a + 7 < 0 \Rightarrow 9a < -7 \Rightarrow a < -\frac{7}{9}$. 7. **Check discriminant for real roots:** $$\Delta = (-4)^2 - 4(a+2)(1) = 16 - 4a - 8 = 8 - 4a.$$ For real roots, $$\Delta \geq 0 \Rightarrow 8 - 4a \geq 0 \Rightarrow a \leq 2.$$ 8. **Combine conditions:** - If $a+2 > 0$ (i.e., $a > -2$) and $a > -\frac{7}{9}$ and $a \leq 2$, then roots are less than 3. - If $a+2 < 0$ (i.e., $a < -2$) and $a < -\frac{7}{9}$ and $a \leq 2$, no $a$ satisfies both $a < -2$ and $a < -\frac{7}{9}$ with real roots because $a < -2$ is stricter. 9. **Final answer:** The roots are less than 3 for $$-\frac{7}{9} < a \leq 2.$$