Stuck on practice exam question: Find the value of α^4 + β^4

The quadratic equation is 2x^2 + 4x + 3 = 0 --> Has roots alpha and beta.
Need to know the formula for alpha^4 + beta^4 in a similar sense to alpha^2 + Beta^2 = (A+B)^2 - 2AB and A^3 + B^3 = (A+B)^3 - 3AB(A+B).
For AS Further Pure Maths 1- Roots of Quadratic Equations.

α + β = -2 ......(1)
αβ = 3/2 => 2αβ = 3......(2)
Square (1): α^2 + 2αβ + β^2 = 4
=> α^2 + β^2 = 4 - 2αβ = 4 - 3 = 1 ......(3)
Square (3): α^4 + 2(αβ)^2 + β^4 = 1
=> α^4 + β^4 = 1 - 2(αβ)^2 = 1 - 2(3/2)^2 = -7/2

Answer: α^4 + β^4 = -7/2

(α^2 + β^2)^2 - 2α^2β^2
= (α + β)^4 - 2α^2β^2 - 2α^2β^2 is incorrect.

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You can derive a formula by treating α^4 + β^4 as (α^2)^2 + (β^2)^2.............

α^4 + β^4
= (α^2)^2 + (β^2)^2
= (α^2 + β^2)^2 - 2α^2β^2
= (α + β)^4 - 2α^2β^2 - 2α^2β^2
= (α + β)^4 - 4α^2β^2

2x^2 + 4x + 3 = 0,
From the quadratic equation, the discriminant = -8
The roots are complex conjugates [- 4 ± √(-8)]/4 = - 1 ± (√(-2))/2 and α + β = -2

Numerically α = -1 + 0.7071i and β = -1 - 0.7071i