If a^4 = 28, then p + 2q = ?

x^2 - x - 5 = 0
Solving by quadratic formula α, β = 1 ± √1+4(2) = 1 ± √5/2
Let α = 1 + √5/2 and β = 1 - √5/2
a^4 = pα^4 + qβ^4
28 = p(1 + √5/2)^4 + q(1 - √5/2)^4
=> 448 = p(1 + 4√5 + 30 + 20√5 + 25) + q(1 - 4√5 + 30 - 20√5 + 25)
=> 448 = p(56 + 24√5) + q(56 - 24√5)
=> 448 = 56p + 56q and 24p√5 - 24q√5 = 0
=> p = q
=> 448 = 56p + 56p
=> 8 = 2p
=> p = 4 = q
∴ p + 2q = 4 + 2(4) = 4 + 8 = 12
Hence option D is the answer.