Find the value
\[ x = \frac{\sqrt{3} + 1}{2} \]
\[ x^2 = \frac{(\sqrt{3} + 1)^2}{4} = \frac{3 + 2\sqrt{3} + 1}{4} = \frac{4 + 2\sqrt{3}}{4} = 1 + \frac{\sqrt{3}}{2} \]
\[ x^3 = x \cdot x^2 = \frac{\sqrt{3} + 1}{2} \left(1 + \frac{\sqrt{3}}{2}\right) \]
\[ = \frac{(\sqrt{3} + 1)(2 + \sqrt{3})}{4} \]
\[ = \frac{2\sqrt{3} + 3 + 2 + \sqrt{3}}{4} = \frac{5 + 3\sqrt{3}}{4} \]
\[ 4x^3 = 5 + 3\sqrt{3} \]
\[ 3x^2 = 3 + \frac{3\sqrt{3}}{2}, \quad -8x = -4\sqrt{3} – 4 \]
\[ \Rightarrow 4x^3 + 3x^2 – 8x + 7 \]
\[ = (5 + 3\sqrt{3}) + \left(3 + \frac{3\sqrt{3}}{2}\right) + (-4\sqrt{3} – 4) + 7 \]
\[ = (5 + 3 – 4 + 7) + \left(3\sqrt{3} + \frac{3\sqrt{3}}{2} – 4\sqrt{3}\right) \]
\[ = 11 + \frac{\sqrt{3}}{2} \]