If `a=(3+sqrt5)/2` then find the vaule of `a^(2)+1/(a^(2))`

Options

A. 5
B. 6
C. 7
D. 8

Explanation

To find the value of \( a^2 + \frac{1}{a^2} \) where \( a = \frac{3 + \sqrt{5}}{2} \), we can follow these steps:

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

First, we need to calculate \( a^2 \):

\[ a^2 = \left( \frac{3 + \sqrt{5}}{2} \right)^2 \]

Using the formula \( (x+y)^2 = x^2 + 2xy + y^2 \):

\[ a^2 = \frac{(3 + \sqrt{5})^2}{2^2} = \frac{9 + 6\sqrt{5} + 5}{4} = \frac{14 + 6\sqrt{5}}{4} = \frac{7 + 3\sqrt{5}}{2} \]

### Step 2: Calculate \( \frac{1}{a} \)

Next, we find \( \frac{1}{a} \):

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

To rationalize the denominator, multiply the numerator and the denominator by \( 3 - \sqrt{5} \):

\[ \frac{1}{a} = \frac{2(3 - \sqrt{5})}{(3 + \sqrt{5})(3 - \sqrt{5})} = \frac{2(3 - \sqrt{5})}{9 - 5} = \frac{2(3 - \sqrt{5})}{4} = \frac{3 - \sqrt{5}}{2} \]

### Step 3: Calculate \( \frac{1}{a^2} \)

Now, we calculate \( \frac{1}{a^2} \):

\[ \frac{1}{a^2} = \left( \frac{1}{a} \right)^2 = \left( \frac{3 - \sqrt{5}}{2} \right)^2 = \frac{(3 - \sqrt{5})^2}{4} \]

Using the formula \( (x-y)^2 = x^2 - 2xy + y^2 \):

\[ \frac{1}{a^2} = \frac{9 - 6\sqrt{5} + 5}{4} = \frac{14 - 6\sqrt{5}}{4} = \frac{7 - 3\sqrt{5}}{2} \]

### Step 4: Calculate \( a^2 + \frac{1}{a^2} \)

Now, we can find \( a^2 + \frac{1}{a^2} \):

\[ a^2 + \frac{1}{a^2} = \frac{7 + 3\sqrt{5}}{2} + \frac{7 - 3\sqrt{5}}{2} \]

Combining the fractions:

\[ a^2 + \frac{1}{a^2} = \frac{(7 + 3\sqrt{5}) + (7 - 3\sqrt{5})}{2} = \frac{14}{2} = 7 \]

### Final Answer

Thus, the value of \( a^2 + \frac{1}{a^2} \) is \( 7 \).

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