If `a^(2)- 3a + 1= 0 and a ne 0`, find : `a^(2) + (1)/(a^(2))`

Answer

To solve the equation \( a^2 - 3a + 1 = 0 \) and find \( a^2 + \frac{1}{a^2} \), we can follow these steps:

### Step 1: Solve for \( a^2 \)
From the equation \( a^2 - 3a + 1 = 0 \), we can express \( a^2 \) in terms of \( a \):
\[ a^2 = 3a - 1 \]

### Step 2: Find \( \frac{1}{a} \)
To find \( \frac{1}{a} \), we can use the quadratic formula. The roots of the equation \( a^2 - 3a + 1 = 0 \) are given by:
\[ a = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{3 \pm \sqrt{(-3)^2 - 4 \cdot 1 \cdot 1}}{2 \cdot 1} = \frac{3 \pm \sqrt{9 - 4}}{2} = \frac{3 \pm \sqrt{5}}{2} \]

### Step 3: Calculate \( a + \frac{1}{a} \)
Using the roots, we can find \( a + \frac{1}{a} \):
\[ \frac{1}{a} = \frac{2}{3 \pm \sqrt{5}} \]
To simplify \( a + \frac{1}{a} \), we can multiply both sides by \( a \):
\[ a + \frac{1}{a} = a + \frac{2}{3 \pm \sqrt{5}} = \frac{a(3 \pm \sqrt{5}) + 2}{3 \pm \sqrt{5}} \]

### Step 4: Square \( a + \frac{1}{a} \)
Now we can square \( a + \frac{1}{a} \):
\[ \left(a + \frac{1}{a}\right)^2 = a^2 + 2 + \frac{1}{a^2} \]
Let \( x = a + \frac{1}{a} \), then:
\[ x^2 = a^2 + 2 + \frac{1}{a^2} \]
Thus,
\[ a^2 + \frac{1}{a^2} = x^2 - 2 \]

### Step 5: Find \( a + \frac{1}{a} \) from \( a^2 - 3a + 1 = 0 \)
Since we know \( a^2 = 3a - 1 \):
\[ x = 3 \]

### Step 6: Substitute \( x \) back to find \( a^2 + \frac{1}{a^2} \)
Now substituting \( x \) back:
\[ a^2 + \frac{1}{a^2} = 3^2 - 2 = 9 - 2 = 7 \]

### Final Answer
Thus, the value of \( a^2 + \frac{1}{a^2} \) is:
\[ \boxed{7} \]