the roots of the equation `(a+sqrt(b))^(x^2-15)+(a-sqrt(b))^(x^2-15)=2a` where `a^2-b=1` are

Options

A. `+-2,+-sqrt(3)`
B. `+-4,+-sqrt(14)`
C. `+-3,+-sqrt(5)`
D. `+-6,+-sqrt(20)`

Explanation

To solve the equation \((a + \sqrt{b})^{x^2 - 15} + (a - \sqrt{b})^{x^2 - 15} = 2a\) given that \(a^2 - b = 1\), we can follow these steps:

### Step 1: Substitute \(b\)
Since we know that \(a^2 - b = 1\), we can express \(b\) in terms of \(a\):
\[ b = a^2 - 1 \]

### Step 2: Rewrite the equation
Substituting \(b\) into the original equation gives us:
\[ (a + \sqrt{a^2 - 1})^{x^2 - 15} + (a - \sqrt{a^2 - 1})^{x^2 - 15} = 2a \]

### Step 3: Let \(y = (a + \sqrt{b})^{x^2 - 15}\)
Let \(y = (a + \sqrt{b})^{x^2 - 15}\). Then, we can rewrite the equation as:
\[ y + \frac{1}{y} = 2a \]

### Step 4: Multiply by \(y\)
Multiplying both sides by \(y\) gives:
\[ y^2 - 2ay + 1 = 0 \]

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

### Step 6: Find the values of \(y\)
Thus, we have:
\[ y = a + \sqrt{a^2 - 1} \quad \text{or} \quad y = a - \sqrt{a^2 - 1} \]

### Step 7: Relate \(y\) back to \(x\)
Since \(y = (a + \sqrt{b})^{x^2 - 15}\), we have two cases to consider:

**Case 1:**
\[ (a + \sqrt{b})^{x^2 - 15} = a + \sqrt{a^2 - 1} \]
Taking logarithms on both sides:
\[ x^2 - 15 = 1 \implies x^2 = 16 \implies x = \pm 4 \]

**Case 2:**
\[ (a + \sqrt{b})^{x^2 - 15} = a - \sqrt{a^2 - 1} \]
Taking logarithms on both sides:
\[ x^2 - 15 = -1 \implies x^2 = 14 \implies x = \pm \sqrt{14} \]

### Final Roots
Thus, the roots of the equation are:
\[ x = 4, -4, \sqrt{14}, -\sqrt{14} \]