If `(5+2sqrt6)^(x^(2)-8)+(5-2sqrt6)^(x^(2)-8)=10, x inR`
On the basis of above information, answer the following questions :
Sum of positive solutions is

Options

A. 3
B. `3+sqrt7`
C. `2+sqrt5`
D. 2

Answer

To solve the equation \( (5 + 2\sqrt{6})^{x^2 - 8} + (5 - 2\sqrt{6})^{x^2 - 8} = 10 \), we can follow these steps:

### Step 1: Let \( y = x^2 - 8 \)
We can rewrite the equation as:
\[ (5 + 2\sqrt{6})^y + (5 - 2\sqrt{6})^y = 10 \]

### Step 2: Recognize the symmetry
Notice that \( 5 - 2\sqrt{6} \) is the reciprocal of \( 5 + 2\sqrt{6} \):
\[ 5 - 2\sqrt{6} = \frac{1}{5 + 2\sqrt{6}} \]
Thus, we can express the equation in terms of \( a = (5 + 2\sqrt{6})^y \):
\[ a + \frac{1}{a} = 10 \]

### Step 3: Multiply through by \( a \)
To eliminate the fraction, we multiply the entire equation by \( a \):
\[ a^2 + 1 = 10a \]

### Step 4: Rearrange into standard quadratic form
Rearranging gives us:
\[ a^2 - 10a + 1 = 0 \]

### Step 5: Solve the quadratic equation
Using the quadratic formula \( a = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
\[ a = \frac{10 \pm \sqrt{(-10)^2 - 4 \cdot 1 \cdot 1}}{2 \cdot 1} = \frac{10 \pm \sqrt{100 - 4}}{2} = \frac{10 \pm \sqrt{96}}{2} = \frac{10 \pm 4\sqrt{6}}{2} = 5 \pm 2\sqrt{6} \]

### Step 6: Find \( y \)
Since \( a = (5 + 2\sqrt{6})^y \), we have two cases:
1. \( (5 + 2\sqrt{6})^y = 5 + 2\sqrt{6} \)
2. \( (5 + 2\sqrt{6})^y = 5 - 2\sqrt{6} \)

For the first case:
\[ y = 1 \implies x^2 - 8 = 1 \implies x^2 = 9 \implies x = 3 \text{ or } x = -3 \]

For the second case:
\[ y = -1 \implies x^2 - 8 = -1 \implies x^2 = 7 \implies x = \sqrt{7} \text{ or } x = -\sqrt{7} \]

### Step 7: Identify positive solutions
The positive solutions are \( x = 3 \) and \( x = \sqrt{7} \).

### Step 8: Calculate the sum of positive solutions
The sum of the positive solutions is:
\[ 3 + \sqrt{7} \]

Thus, the final answer is:
\[ \text{Sum of positive solutions} = 3 + \sqrt{7} \]

---