If `(5+2sqrt6)^(x^(2)-3)+(5-2sqrt6)^(x^(2)-3)=10`, then x =

Options

A. 2,2
B. `sqrt2,-sqrt2`
C. `2,+sqrt2`
D. `2,-2,sqrt2,-sqrt2`

Answer

To solve the equation \((5 + 2\sqrt{6})^{x^2 - 3} + (5 - 2\sqrt{6})^{x^2 - 3} = 10\), we can follow these steps:

### Step 1: Set a substitution
Let \(y = x^2 - 3\). Then, the equation becomes:
\[ (5 + 2\sqrt{6})^y + (5 - 2\sqrt{6})^y = 10 \]

### Step 2: Analyze the terms
Notice that \(5 + 2\sqrt{6}\) and \(5 - 2\sqrt{6}\) are conjugates. Let's denote:
\[ a = 5 + 2\sqrt{6} \quad \text{and} \quad b = 5 - 2\sqrt{6} \]

### Step 3: Find the product of \(a\) and \(b\)
Calculate \(ab\):
\[ ab = (5 + 2\sqrt{6})(5 - 2\sqrt{6}) = 5^2 - (2\sqrt{6})^2 = 25 - 24 = 1 \]

### Step 4: Use properties of exponents
Since \(ab = 1\), we have:
\[ b = \frac{1}{a} \]
Thus, we can rewrite the equation as:
\[ a^y + \left(\frac{1}{a}\right)^y = 10 \]
This simplifies to:
\[ a^y + \frac{1}{a^y} = 10 \]

### Step 5: Set \(z = a^y\)
Now, let \(z = a^y\). The equation becomes:
\[ z + \frac{1}{z} = 10 \]

### Step 6: Multiply through by \(z\)
Multiply both sides by \(z\) (assuming \(z \neq 0\)):
\[ z^2 + 1 = 10z \]
Rearranging gives:
\[ z^2 - 10z + 1 = 0 \]

### Step 7: Solve the quadratic equation
Using the quadratic formula:
\[ z = \frac{10 \pm \sqrt{10^2 - 4 \cdot 1 \cdot 1}}{2 \cdot 1} = \frac{10 \pm \sqrt{100 - 4}}{2} = \frac{10 \pm \sqrt{96}}{2} = \frac{10 \pm 4\sqrt{6}}{2} = 5 \pm 2\sqrt{6} \]

### Step 8: Find \(y\)
We have two cases for \(z\):
1. \(z = 5 + 2\sqrt{6}\)
2. \(z = 5 - 2\sqrt{6}\)

Since \(z = a^y\), we can take the logarithm:
1. For \(z = 5 + 2\sqrt{6}\):
\[ a^y = 5 + 2\sqrt{6} \implies y = 1 \]
2. For \(z = 5 - 2\sqrt{6}\):
\[ a^y = 5 - 2\sqrt{6} \implies y = -1 \]

### Step 9: Substitute back for \(y\)
Recall \(y = x^2 - 3\):
1. If \(y = 1\):
\[ x^2 - 3 = 1 \implies x^2 = 4 \implies x = \pm 2 \]
2. If \(y = -1\):
\[ x^2 - 3 = -1 \implies x^2 = 2 \implies x = \pm \sqrt{2} \]

### Step 10: Final solutions
Thus, the values of \(x\) are:
\[ x = 2, -2, \sqrt{2}, -\sqrt{2} \]