If `a= sqrt((sqrt5+1)/(sqrt5-1))` then `sqrt(a^2-a-1)` is equal to _____.

Options

A. A) 5
B. B) `sqrt5`
C. C) 0
D. D) 1

Answer

To solve the problem, we start with the expression given for \( a \):

\[ a = \sqrt{\frac{\sqrt{5} + 1}{\sqrt{5} - 1}} \]

### Step 1: Rationalize the denominator

To simplify \( a \), we will rationalize the denominator by multiplying the numerator and denominator by the conjugate of the denominator, which is \( \sqrt{5} + 1 \):

\[ a = \sqrt{\frac{(\sqrt{5} + 1)(\sqrt{5} + 1)}{(\sqrt{5} - 1)(\sqrt{5} + 1)}} \]

### Step 2: Simplify the numerator and denominator

Calculating the numerator:

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

Calculating the denominator using the difference of squares:

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

Thus, we have:

\[ a = \sqrt{\frac{6 + 2\sqrt{5}}{4}} = \sqrt{\frac{6 + 2\sqrt{5}}{4}} = \frac{\sqrt{6 + 2\sqrt{5}}}{2} \]

### Step 3: Find \( a^2 \)

Now we calculate \( a^2 \):

\[ a^2 = \frac{6 + 2\sqrt{5}}{4} = \frac{3 + \sqrt{5}}{2} \]

### Step 4: Calculate \( a^2 - a - 1 \)

Now we need to compute \( a^2 - a - 1 \):

\[ a^2 - a - 1 = \frac{3 + \sqrt{5}}{2} - \frac{1}{2} - 1 \]

Finding a common denominator:

\[ = \frac{3 + \sqrt{5}}{2} - \frac{1}{2} - \frac{2}{2} = \frac{3 + \sqrt{5} - 1 - 2}{2} = \frac{\sqrt{5}}{2} \]

### Step 5: Calculate \( \sqrt{a^2 - a - 1} \)

Now we find \( \sqrt{a^2 - a - 1} \):

\[ \sqrt{a^2 - a - 1} = \sqrt{\frac{\sqrt{5}}{2}} = \frac{\sqrt{\sqrt{5}}}{\sqrt{2}} = \frac{5^{1/4}}{\sqrt{2}} \]

### Final Answer

The final answer for \( \sqrt{a^2 - a - 1} \) is:

\[ \frac{5^{1/4}}{\sqrt{2}} \]