1. **Stating the problem:** We are given the quadratic equation $$16x^2 - 8x - 47 = 0$$ and told its roots are $$\frac{1}{4} + \sqrt{5}$$ and $$\frac{1}{4} - \sqrt{5}$$. We need to find the value of $$b$$ where $$b > 0$$ in the root expression $$\frac{1}{4} \pm b$$. 2. **Recall the quadratic formula:** For a quadratic equation $$ax^2 + bx + c = 0$$, the roots are given by $$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$. 3. **Identify coefficients:** From the equation, $$a = 16$$, $$b = -8$$, and $$c = -47$$. 4. **Calculate the roots using the quadratic formula:** $$x = \frac{-(-8) \pm \sqrt{(-8)^2 - 4 \times 16 \times (-47)}}{2 \times 16} = \frac{8 \pm \sqrt{64 + 3008}}{32} = \frac{8 \pm \sqrt{3072}}{32}$$ 5. **Simplify the square root:** $$\sqrt{3072} = \sqrt{1024 \times 3} = 32 \sqrt{3}$$ 6. **Rewrite the roots:** $$x = \frac{8 \pm 32 \sqrt{3}}{32} = \frac{8}{32} \pm \frac{32 \sqrt{3}}{32} = \frac{1}{4} \pm \sqrt{3}$$ 7. **Compare with given roots:** The problem states roots are $$\frac{1}{4} \pm \sqrt{5}$$ but our calculation shows roots are $$\frac{1}{4} \pm \sqrt{3}$$. 8. **Conclusion:** The value of $$b$$ in the root expression $$\frac{1}{4} \pm b$$ is $$\sqrt{3}$$, which is positive. **Final answer:** $$b = \sqrt{3}$$