Let `p ,q` be integers and let `alpha,beta` be the roots of the equation, `x^2-x-1=0,` where `alpha!=beta` . For `n=0,1,2, ,l e ta_n=palpha^n+qbeta^ndot` FACT : If `aa n db` are rational number and `a+bsqrt(5)=0,t h e na=0=bdot` If `a_4=28 ,t h e np+2q=` 7 (b) 21 (c) 14 (d) 12

Options

A. 21
B. 14
C. 7
D. 12

Explanation

To solve the problem step by step, we start by analyzing the given quadratic equation and the roots.

### Step 1: Identify the roots of the equation
The equation given is:
\[ x^2 - x - 1 = 0 \]
Using the quadratic formula, the roots \( \alpha \) and \( \beta \) can be calculated as:
\[ \alpha, \beta = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
where \( a = 1, b = -1, c = -1 \).

Calculating the discriminant:
\[ b^2 - 4ac = (-1)^2 - 4 \cdot 1 \cdot (-1) = 1 + 4 = 5 \]

Now substituting into the formula:
\[ \alpha, \beta = \frac{1 \pm \sqrt{5}}{2} \]

Thus, we have:
\[ \alpha = \frac{1 + \sqrt{5}}{2}, \quad \beta = \frac{1 - \sqrt{5}}{2} \]

### Step 2: Write the expression for \( a_n \)
The expression for \( a_n \) is given as:
\[ a_n = p \alpha^n + q \beta^n \]
where \( p \) and \( q \) are integers.

### Step 3: Find \( a_4 \)
Using the recurrence relation derived from the roots of the quadratic equation:
\[ a_n = a_{n-1} + a_{n-2} \]
We can express \( a_4 \) in terms of previous terms:
\[ a_4 = a_3 + a_2 \]

Continuing this, we express \( a_3 \) and \( a_2 \):
\[ a_3 = a_2 + a_1 \]
\[ a_2 = a_1 + a_0 \]

Thus, we can express \( a_4 \) as:
\[ a_4 = (a_2 + a_1) + a_2 = 2a_2 + a_1 \]

Now substituting \( a_2 \):
\[ a_4 = 2(a_1 + a_0) + a_1 = 3a_1 + 2a_0 \]

### Step 4: Substitute known values
We know \( a_0 = p + q \) and \( a_1 = p \alpha + q \beta \). Thus:
\[ a_4 = 3(p \alpha + q \beta) + 2(p + q) \]

### Step 5: Use the given value of \( a_4 \)
We are given that \( a_4 = 28 \):
\[ 3(p \alpha + q \beta) + 2(p + q) = 28 \]

### Step 6: Simplify the equation
Using the fact that \( \alpha + \beta = 1 \) and \( \alpha \beta = -1 \):
\[ p \alpha + q \beta = p \alpha + q(1 - \alpha) = (p - q) \alpha + q \]

Substituting this back into the equation:
\[ 3((p - q) \alpha + q) + 2(p + q) = 28 \]
This simplifies to:
\[ 3(p - q) \alpha + 3q + 2p + 2q = 28 \]
\[ 3(p - q) \alpha + 5q + 2p = 28 \]

### Step 7: Solve for \( p \) and \( q \)
Since \( \alpha \) is irrational, the coefficient of \( \alpha \) must equal zero:
\[ 3(p - q) = 0 \Rightarrow p = q \]

Substituting \( p = q \) into the equation:
\[ 5p + 2p = 28 \]
\[ 7p = 28 \]
\[ p = 4 \]

Thus, \( q = 4 \).

### Step 8: Find \( p + 2q \)
Now, we can find \( p + 2q \):
\[ p + 2q = 4 + 2 \cdot 4 = 4 + 8 = 12 \]

### Final Answer
The value of \( p + 2q \) is:
\[ \boxed{12} \]