Simplifying the Radical Expression Value

The question asks for the value of the expression $\frac{\sqrt{5}+1}{\sqrt{5}-1}$. To simplify this, we rationalize the denominator.

Rationalizing the Denominator

The conjugate of the denominator $(\sqrt{5}-1)$ is $(\sqrt{5}+1)$. We multiply both the numerator and the denominator by this conjugate:

$ \frac{\sqrt{5}+1}{\sqrt{5}-1} \times \frac{\sqrt{5}+1}{\sqrt{5}+1} $

Calculating the New Numerator

The new numerator is $(\sqrt{5}+1)(\sqrt{5}+1)$, which is $(\sqrt{5}+1)^2$. Using the algebraic identity $(a+b)^2 = a^2 + 2ab + b^2$:

$ (\sqrt{5}+1)^2 = (\sqrt{5})^2 + 2(\sqrt{5})(1) + (1)^2 = 5 + 2\sqrt{5} + 1 = 6 + 2\sqrt{5} $

Calculating the New Denominator

The new denominator is $(\sqrt{5}-1)(\sqrt{5}+1)$. Using the algebraic identity $(a-b)(a+b) = a^2 - b^2$:

$ (\sqrt{5}-1)(\sqrt{5}+1) = (\sqrt{5})^2 - (1)^2 = 5 - 1 = 4 $

Final Value Calculation

Substitute the calculated numerator and denominator back:

$ \frac{6 + 2\sqrt{5}}{4} $

Simplify the fraction by factoring out 2 from the numerator and cancelling:

$ \frac{2(3 + \sqrt{5})}{4} = \frac{3 + \sqrt{5}}{2} $

The value of the expression is $\frac{3+\sqrt{5}}{2}$.