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. `pm 4`
B. `pm3`
C. `pm sqrt(14)`
D. `pm sqrt (5)`

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 \(a^2 - b = 1\)
We know that \(a^2 - b = 1\). This can be rewritten as:
\[ b = a^2 - 1 \]

### Step 2: Rewrite the equation
Using the identity \(a^2 - b = (a + \sqrt{b})(a - \sqrt{b})\), we can express \(a + \sqrt{b}\) in terms of \(a - \sqrt{b}\):
\[ a + \sqrt{b} = \frac{1}{a - \sqrt{b}} \]
Now, substitute this into the original equation:
\[ \left(\frac{1}{a - \sqrt{b}}\right)^{x^2 - 15} + (a - \sqrt{b})^{x^2 - 15} = 2a \]

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

### Step 4: Multiply through by \(y\)
Multiply the entire equation by \(y\) to eliminate the fraction:
\[ 1 + y^2 = 2ay \]

### Step 5: Rearrange into a quadratic equation
Rearranging gives us:
\[ y^2 - 2ay + 1 = 0 \]

### Step 6: Use the quadratic formula
Using the quadratic formula \(y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\):
\[ y = \frac{2a \pm \sqrt{(2a)^2 - 4 \cdot 1 \cdot 1}}{2 \cdot 1} \]
\[ y = \frac{2a \pm \sqrt{4a^2 - 4}}{2} \]
\[ y = a \pm \sqrt{a^2 - 1} \]

### Step 7: Substitute back for \(y\)
Recall that \(y = (a - \sqrt{b})^{x^2 - 15}\). Thus, we have:
\[ (a - \sqrt{b})^{x^2 - 15} = a \pm \sqrt{a^2 - 1} \]

### Step 8: Solve for \(x^2 - 15\)
Taking logarithms on both sides, we can solve for \(x^2 - 15\):
1. For \(y = a + \sqrt{a^2 - 1}\):
\[ x^2 - 15 = 1 \implies x^2 = 16 \implies x = \pm 4 \]
2. For \(y = a - \sqrt{a^2 - 1}\):
\[ x^2 - 15 = -1 \implies x^2 = 14 \implies x = \pm \sqrt{14} \]

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