Explanation:
Okay, so I need to find the unordered set {α₁α₂ + α₃α₄, α₁α₃ + α₂α₄, α₁α₄ + α₂α₃} where α₁, α₂, α₃, α₄ are the roots of the equation x⁴ + 2x³ + 2 = 0. Hmm, let me recall Vieta's formulas for quartic equations.

First, for a quartic equation x⁴ + a x³ + b x² + c x + d = 0, the sums of roots are related to the coefficients. Specifically, if the roots are α, β, γ, δ, then:

α + β + γ + δ = -a

αβ + αγ + αδ + βγ + βδ + γδ = b

αβγ + αβδ + αγδ + βγδ = -c

αβγδ = d

In my given equation, the equation is x⁴ + 2x³ + 0x² + 0x + 2 = 0. Wait, so the coefficients are:

a = 2, b = 0, c = 0, d = 2.

So applying Vieta's formulas:

Sum of roots: α₁ + α₂ + α₃ + α₄ = -a = -2

Sum of products two at a time: α₁α₂ + α₁α₃ + α₁α₄ + α₂α₃ + α₂α₄ + α₃α₄ = b = 0

Sum of products three at a time: α₁α₂α₃ + α₁α₂α₄ + α₁α₃α₄ + α₂α₃α₄ = -c = 0

Product of roots: α₁α₂α₃α₄ = d = 2

Okay, so we need to compute the three expressions:

S1 = α₁α₂ + α₃α₄

S2 = α₁α₃ + α₂α₄

S3 = α₁α₄ + α₂α₃

And then put them in an unordered set {S1, S2, S3}

So, the question is to find these three values.

Now, note that S1 + S2 + S3 = (α₁α₂ + α₃α₄) + (α₁α₃ + α₂α₄) + (α₁α₄ + α₂α₃) = sum of all products two at a time, which is given by Vieta as 0. Therefore, S1 + S2 + S3 = 0.

So the three numbers add up to zero. But we need to find each of them.

But how can we find individual values?

Alternatively, maybe we can think of them as the sums of products of pairs of roots, grouped into two pairs.

Let me think. Suppose we consider that the roots are grouped into pairs. For example, S1 is the sum of products of two pairs. However, since there are three such expressions, each corresponding to a different way of grouping the roots into two pairs.

But perhaps I can relate these expressions to symmetric sums?

Alternatively, maybe I can consider variables. Let me denote:

Let me think of the roots as α, β, γ, δ. Let me denote the three expressions as:

S1 = αβ + γδ

S2 = αγ + βδ

S3 = αδ + βγ

So, if we can find S1, S2, S3, then we can put them in the set.

But how?

Since we know that S1 + S2 + S3 = 0. Also, maybe we can compute S1*S2 + S1*S3 + S2*S3? Or perhaps other combinations? Alternatively, maybe we can find another equation involving S1, S2, S3.

Alternatively, perhaps consider that if we can find expressions for S1, S2, S3 in terms of the known Vieta's formulas.

Alternatively, perhaps consider that if we take the product (x - α)(x - β)(x - γ)(x - δ) = x⁴ + 2x³ + 2.

But maybe another approach: Let me think about the roots. Let me suppose that the roots come in pairs? Not sure.

Alternatively, perhaps consider that if we can find S1, S2, S3, maybe we can relate them through some equations. Let me try:

Let me consider that:

Let me denote S = S1 + S2 + S3 = 0.

Let me denote P = product of roots = αβγδ = 2.

Also, the sum of products two at a time is zero. Let me see:

Also, let me think about the sum S1*S2 + S1*S3 + S2*S3?

But that seems complicated. Alternatively, maybe think about individual S1, S2, S3. Let me try squaring the sum of roots, for example.

We know that (α + β + γ + δ)² = α² + β² + γ² + δ² + 2(αβ + αγ + αδ + βγ + βδ + γδ)

But we know that α + β + γ + δ = -2, so (-2)² = 4 = (α² + β² + γ² + δ²) + 2*(0) => α² + β² + γ² + δ² = 4.

But not sure if that helps yet.

Alternatively, maybe think about the values S1, S2, S3. Let me take S1 = αβ + γδ. Maybe if I can find expressions for αβ, γδ, etc.?

Alternatively, let me think of S1, S2, S3 as the sums of two different products. Let me think about variables. Let me suppose that:

Let me denote P1 = αβ, P2 = γδ. Then S1 = P1 + P2.

Similarly, for S2 and S3, but they would involve different pairings.

Alternatively, perhaps consider that if I can find P1, P2, etc. But maybe not.

Alternatively, think about the equations involving Vieta's formulas. Let me think. Let me also note that:

From Vieta's formula, we know that:

αβ + αγ + αδ + βγ + βδ + γδ = 0.

But also, if we take S1 = αβ + γδ, then S2 = αγ + βδ, and S3 = αδ + βγ.

So, if we add S1 + S2 + S3 = 0 as before. Now, if I can find other relations between S1, S2, S3.

Let me also note that:

Suppose we take S1 = αβ + γδ. If I multiply S1 by something?

Alternatively, consider that maybe S1 * something?

Alternatively, think about the product of the roots. Since αβγδ = 2, so (αβ)(γδ) = 2. Therefore, if S1 = αβ + γδ, then S1 is the sum of two numbers whose product is 2. Similarly for S2 and S3.

Wait, but S2 = αγ + βδ. Let me check if (αγ)(βδ) = αβγδ = 2. Similarly for S3: (αδ)(βγ) = αβγδ = 2. So each of the S1, S2, S3 is the sum of two numbers whose product is 2. Interesting!

Therefore, each S1, S2, S3 satisfies the equation t² - (sum)t + 2 = 0. Wait, but each S_i is a sum of two numbers whose product is 2. So, for example, S1 = p + q, where pq = 2. Therefore, p and q are roots of the equation x² - S1 x + 2 = 0.

But maybe this is not directly helpful? Hmm.

But perhaps if we can find S1, S2, S3. Let me consider that S1, S2, S3 are roots of some cubic equation? Maybe?

Alternatively, since we have S1 + S2 + S3 = 0, and also perhaps S1S2 + S1S3 + S2S3 and S1S2S3?

If I can find these two other symmetric sums, then I can find the set {S1, S2, S3}.

So let me try to compute S1S2 + S1S3 + S2S3 and S1S2S3.

First, let me compute S1S2 + S1S3 + S2S3.

Alternatively, note that (S1 + S2 + S3)^2 = S1² + S2² + S3² + 2(S1S2 + S1S3 + S2S3). Since S1 + S2 + S3 = 0, this implies that 0 = S1² + S2² + S3² + 2(S1S2 + S1S3 + S2S3), so S1² + S2² + S3² = -2(S1S2 + S1S3 + S2S3). But not sure if helpful yet.

Alternatively, perhaps compute S1S2 + S1S3 + S2S3 directly. Let me see.

Let me expand S1S2:

S1*S2 = (αβ + γδ)(αγ + βδ)

Let me compute this:

First, expand the product:

= αβ*αγ + αβ*βδ + γδ*αγ + γδ*βδ

= α²βγ + αβ²δ + αγ²δ + βγδ²

Similarly, S1*S3 = (αβ + γδ)(αδ + βγ)

= αβ*αδ + αβ*βγ + γδ*αδ + γδ*βγ

= α²βδ + αβ²γ + αδ²γ + βγ²δ

Similarly, S2*S3 = (αγ + βδ)(αδ + βγ)

= αγ*αδ + αγ*βγ + βδ*αδ + βδ*βγ

= α²γδ + αβγ² + αβδ² + β²γδ

Therefore, adding S1*S2 + S1*S3 + S2*S3:

= [α²βγ + αβ²δ + αγ²δ + βγδ²] + [α²βδ + αβ²γ + αδ²γ + βγ²δ] + [α²γδ + αβγ² + αβδ² + β²γδ]

Hmm, this seems complicated, but maybe we can combine like terms.

Let me list all terms:

From S1*S2:

Term1: α²βγ

Term2: αβ²δ

Term3: αγ²δ

Term4: βγδ²

From S1*S3:

Term5: α²βδ

Term6: αβ²γ

Term7: αδ²γ

Term8: βγ²δ

From S2*S3:

Term9: α²γδ

Term10: αβγ²

Term11: αβδ²

Term12: β²γδ

Now, let me see if these terms can be grouped together.

For example, Term1: α²βγ and Term6: αβ²γ. These have αβγ as a common factor.

Similarly, Term2: αβ²δ and Term5: α²βδ have αβδ as a common factor.

Similarly, Term3: αγ²δ and Term7: αδ²γ have αγδ as common factor.

Term4: βγδ² and Term12: β²γδ have βγδ as common factor.

Term8: βγ²δ and Term10: αβγ² have βγ² as common factor?

Wait, Term8 is βγ²δ and Term10 is αβγ². So they share βγ². Similarly Term11: αβδ² and Term4? Not sure.

Alternatively, maybe factor variables.

Alternatively, maybe factor out common terms. Let me see:

Let me consider each term:

Term1: α²βγ

Term2: αβ²δ

Term3: αγ²δ

Term4: βγδ²

Term5: α²βδ

Term6: αβ²γ

Term7: αγδ²

Term8: βγ²δ

Term9: α²γδ

Term10: αβγ²

Term11: αβδ²

Term12: β²γδ

So, maybe group terms with similar variables.

For example, terms involving α²βγ, α²βδ, α²γδ, and similar.

Let me group terms with α²:

- α²βγ (Term1)

- α²βδ (Term5)

- α²γδ (Term9)

Similarly for β²:

- αβ²δ (Term2)

- αβ²γ (Term6)

- β²γδ (Term12)

For γ²:

- αγ²δ (Term3)

- βγ²δ (Term8)

- αβγ² (Term10)

For δ²:

- βγδ² (Term4)

- αγδ² (Term7)

- αβδ² (Term11)

Wait, so each of these groups has three terms, and each group corresponds to variables squared. So, for example, the α² terms are α²βγ, α²βδ, α²γδ. Similarly for others.

So, let me compute the sum:

Sum = [α²βγ + α²βδ + α²γδ] + [αβ²δ + αβ²γ + β²γδ] + [αγ²δ + βγ²δ + αβγ²] + [βγδ² + αγδ² + αβδ²]

Wait, but actually, each variable squared has three terms. However, in the previous list, each of the four terms (α², β², γ², δ²) have three terms? Wait, no. Wait, looking at the above:

- The α² terms: Term1, Term5, Term9

- The β² terms: Term2, Term6, Term12

- The γ² terms: Term3, Term8, Term10

- The δ² terms: Term4, Term7, Term11

So yes, each squared variable has three terms. Therefore, the entire sum is:

Sum = (α²βγ + α²βδ + α²γδ) + (αβ²δ + αβ²γ + β²γδ) + (αγ²δ + βγ²δ + αβγ²) + (βγδ² + αγδ² + αβδ²)

But note that each of these groups is the same as:

For example, the first group: α²βγ + α²βδ + α²γδ = α²β(γ + δ) + α²γδ. Similarly for others. But perhaps there is a better way to express this.

Alternatively, note that each group is similar to the product of α² times the sum of products of β, γ, δ?

Wait, not sure. Alternatively, factor terms.

Alternatively, notice that each term in the sum is of the form (variable squared)*(product of two others). For example, α²βγ is α² times βγ. Similarly for all others.

So, the entire sum can be written as:

Sum = α²βγ + α²βδ + α²γδ + αβ²δ + αβ²γ + β²γδ + αγ²δ + βγ²δ + αβγ² + βγδ² + αγδ² + αβδ²

Which can be grouped as:

= α²(βγ + βδ + γδ) + β²(αδ + αγ + γδ) + γ²(αδ + αβ + βδ) + δ²(αβ + αγ + βγ)

But wait, this seems complicated. Wait, let me check:

For instance, the first term α²(βγ + βδ + γδ). Is that true? Let me see:

α²βγ + α²βδ + α²γδ = α²(βγ + βδ + γδ). Yes. Similarly for the others.

So, the Sum is equal to:

α²(βγ + βδ + γδ) + β²(αδ + αγ + γδ) + γ²(αδ + αβ + βδ) + δ²(αβ + αγ + βγ)

But note that βγ + βδ + γδ is equal to sum of products of β, γ, δ. Let me recall Vieta's formula.

From Vieta's, we know that:

α + β + γ + δ = -2

αβ + αγ + αδ + βγ + βδ + γδ = 0

αβγ + αβδ + αγδ + βγδ = 0

αβγδ = 2

So, if we take βγ + βδ + γδ, this is equal to [αβ + αγ + αδ + βγ + βδ + γδ] - α(β + γ + δ) = 0 - α(β + γ + δ)

Similarly, since α + β + γ + δ = -2, then β + γ + δ = -2 - α. Therefore:

βγ + βδ + γδ = -α(-2 - α) = 2α + α².

Similarly, for the other terms:

For example, αδ + αγ + γδ = -β(-2 - β) = 2β + β²

Similarly, αδ + αβ + βδ = -γ(-2 - γ) = 2γ + γ²

And αβ + αγ + βγ = -δ(-2 - δ) = 2δ + δ²

Therefore, substituting back into the Sum:

Sum = α²(2α + α²) + β²(2β + β²) + γ²(2γ + γ²) + δ²(2δ + δ²)

= 2α³ + α⁴ + 2β³ + β⁴ + 2γ³ + γ⁴ + 2δ³ + δ⁴

= (α⁴ + β⁴ + γ⁴ + δ⁴) + 2(α³ + β³ + γ³ + δ³)

So, we have that S1*S2 + S1*S3 + S2*S3 equals this expression. Therefore, if we can compute α⁴ + β⁴ + γ⁴ + δ⁴ and α³ + β³ + γ³ + δ³, then we can compute this Sum.

Okay, so let's compute α³ + β³ + γ³ + δ³.

There's a formula for the sum of cubes. Recall that:

α³ + β³ + γ³ + δ³ = (α + β + γ + δ)³ - 3(α + β + γ + δ)(αβ + αγ + αδ + βγ + βδ + γδ) + 3(αβγ + αβδ + αγδ + βγδ)

But let me check that formula. For three variables, the formula is a³ + b³ + c³ = (a + b + c)^3 - 3(a + b + c)(ab + bc + ac) + 3abc. For four variables, perhaps there is an analogous formula. Alternatively, maybe use Newton's identities?

Alternatively, since we know the values of the power sums. Let me recall that power sums can be computed using recurrence relations.

Alternatively, since the roots satisfy the equation x⁴ + 2x³ + 2 = 0, so for each root, x⁴ = -2x³ - 2. Therefore, for each root, x⁴ = -2x³ - 2. Therefore, we can use this to express higher powers in terms of lower powers.

Therefore, we can compute α³ + β³ + γ³ + δ³ and α⁴ + β⁴ + γ⁴ + δ⁴.

Let me first compute α³ + β³ + γ³ + δ³.

Let me denote S1 = α + β + γ + δ = -2

S2 = αβ + αγ + αδ + βγ + βδ + γδ = 0

S3 = αβγ + αβδ + αγδ + βγδ = 0

S4 = αβγδ = 2

Let me recall that the sum of cubes can be calculated by:

Σα³ = (Σα)^3 - 3ΣαΣβγ + 3Σαβγ

Wait, for four variables, the formula is:

Σα³ = (Σα)^3 - 3ΣαΣβγ + 3Σαβγ

Wait, actually, let me check:

For n variables, Σα³ can be calculated using the formula:

Σα³ = (Σα)^3 - 3ΣαΣβγ + 3Σαβγ. Hmm, not sure. Let me check with three variables.

For three variables, the formula is a³ + b³ + c³ = (a + b + c)^3 - 3(a + b + c)(ab + bc + ac) + 3abc. So yes, for four variables, perhaps:

Σα³ = (Σα)^3 - 3Σα*(Σβγ) + 3Σαβγ. Let me check if this holds.

Let me consider expanding (Σα)^3:

(Σα)^3 = Σα³ + 3Σα²β + 6Σαβγ

Wait, no. Let me recall that:

(Σα)^3 = Σα³ + 3Σα²β + 6αβγ. Wait, for four variables, it's more complicated. Let me think.

Alternatively, maybe use Newton's identities, which relate power sums to elementary symmetric sums.

Newton's identities relate the power sums P_k = α^k + β^k + γ^k + δ^k with the elementary symmetric polynomials.

For a quartic equation, Newton's identities can be used. Let me recall that.

For k ≥ 1,

P_k + a P_{k-1} + b P_{k-2} + c P_{k-3} + d P_{k-4} = 0

Wait, where the equation is x^4 + a x^3 + b x^2 + c x + d = 0. But in our case, the equation is x⁴ + 2x³ + 0x² + 0x + 2 = 0. Therefore, a=2, b=0, c=0, d=2.

Therefore, Newton's identities state that:

For k ≥ 1,

P_k + 2 P_{k-1} + 0*P_{k-2} + 0*P_{k-3} + 2 P_{k-4} = 0.

But note that for k < 4, P_{k-4} would be zero? Wait, maybe not. Wait, Newton's identities for monic polynomials:

If the polynomial is x^4 + a x^3 + b x^2 + c x + d, then:

P_1 + a = 0

P_2 + a P_1 + 2b = 0

P_3 + a P_2 + b P_1 + 3c = 0

P_4 + a P_3 + b P_2 + c P_1 + 4d = 0

And for k > 4:

P_k + a P_{k-1} + b P_{k-2} + c P_{k-3} + d P_{k-4} = 0

Therefore, let me compute P_1, P_2, P_3, P_4.

Given that:

a = 2, b = 0, c = 0, d = 2.

First, P_1:

From the first identity: P_1 + a = 0 ⇒ P_1 = -a = -2. Which matches our earlier Vieta's formula.

Next, P_2:

Second identity: P_2 + a P_1 + 2b = 0 ⇒ P_2 + 2*(-2) + 0 = 0 ⇒ P_2 - 4 = 0 ⇒ P_2 = 4.

Wait, but earlier I calculated that α² + β² + γ² + δ² = 4? From the square of the sum. Yes, because (α + β + γ + δ)^2 = 4 = sum of squares + 2*(sum of products). But sum of products is 0, so sum of squares = 4. Which matches P_2 = 4. Good.

Next, P_3:

Third identity: P_3 + a P_2 + b P_1 + 3c = 0 ⇒ P_3 + 2*4 + 0*(-2) + 0 = 0 ⇒ P_3 + 8 = 0 ⇒ P_3 = -8.

So the sum of cubes is -8.

Now, P_4:

Fourth identity: P_4 + a P_3 + b P_2 + c P_1 + 4d = 0 ⇒ P_4 + 2*(-8) + 0*4 + 0*(-2) + 4*2 = 0 ⇒ P_4 - 16 + 8 = 0 ⇒ P_4 - 8 = 0 ⇒ P_4 = 8.

Alternatively, since each root satisfies x⁴ = -2x³ -2, then P_4 = Σx⁴ = Σ(-2x³ -2) = -2Σx³ - 2*4 = -2*(-8) - 8 = 16 - 8 = 8. Which matches. Good.

Now, going back to our earlier expression for S1*S2 + S1*S3 + S2*S3 = (α⁴ + β⁴ + γ⁴ + δ⁴) + 2(α³ + β³ + γ³ + δ³) = P_4 + 2P_3 = 8 + 2*(-8) = 8 - 16 = -8.

So S1*S2 + S1*S3 + S2*S3 = -8.

Earlier, we had S1 + S2 + S3 = 0.

Therefore, we have the following for the symmetric sums of S1, S2, S3:

Let me denote S = S1 + S2 + S3 = 0

Q = S1*S2 + S1*S3 + S2*S3 = -8

Now, if I can find R = S1*S2*S3, then I can form the cubic equation whose roots are S1, S2, S3:

x³ - Sx² + Qx - R = 0 ⇒ x³ + 0x² -8x - R = 0 ⇒ x³ -8x - R = 0.

So if we can find R, then we can find the cubic equation, and then find its roots.

So, need to compute R = S1*S2*S3.

How?

Let me think. Let me note that S1 = αβ + γδ, S2 = αγ + βδ, S3 = αδ + βγ.

So, S1*S2*S3 = (αβ + γδ)(αγ + βδ)(αδ + βγ)

This seems complicated, but maybe we can expand it?

Alternatively, perhaps use relations from Vieta's formulas.

Alternatively, note that if we consider the product (αβ + γδ)(αγ + βδ)(αδ + βγ). Maybe there's a clever way to compute this.

Alternatively, think about the product (αβ + γδ)(αγ + βδ)(αδ + βγ). Let me first compute (αβ + γδ)(αγ + βδ).

Earlier, we had expanded this as α²βγ + αβ²δ + αγ²δ + βγδ² + ... Wait, no, earlier expansion for S1*S2 gave terms that led us to P_4 + 2P_3. Wait, no, but here we need to compute S1*S2*S3, which is multiplying three such terms.

Alternatively, perhaps use substitution variables.

Alternatively, think that since αβγδ = 2, so perhaps set variables such that:

Let me consider that for example, let me set:

Let me denote p = αβ, q = γδ, so S1 = p + q. Similarly, for S2 and S3, but need to relate them to p and q.

But S2 = αγ + βδ. Hmm. Let me see:

Let me note that αγ + βδ. If p = αβ, q = γδ, then maybe express S2 in terms of p and q?

Alternatively, perhaps consider that αγ + βδ. Let me think. If we consider product αγ * βδ = (αβ)(γδ) = pq = 2. So similar to S1.

But maybe not helpful yet.

Alternatively, think of αγ + βδ as a combination of terms.

Alternatively, maybe use the fact that α, β, γ, δ are roots of the equation, so we can relate them through symmetric sums.

Alternatively, perhaps note that the product S1*S2*S3 can be expressed in terms of known symmetric sums.

Alternatively, let me try to compute S1*S2*S3.

Let me consider that:

First, compute (αβ + γδ)(αγ + βδ) = αβαγ + αββδ + γδαγ + γδβδ = α²βγ + αβ²δ + αγ²δ + βγδ². Which we had earlier. Then multiply this by (αδ + βγ):

So, multiplying the above by (αδ + βγ):

= [α²βγ + αβ²δ + αγ²δ + βγδ²] * [αδ + βγ]

This is going to be a lot, but let me try.

First, expand term by term:

First term: α²βγ * αδ = α³βγδ

Second term: α²βγ * βγ = α²β²γ²

Third term: αβ²δ * αδ = α²β²δ²

Fourth term: αβ²δ * βγ = αβ³γδ

Fifth term: αγ²δ * αδ = α²γ²δ²

Sixth term: αγ²δ * βγ = αβγ³δ

Seventh term: βγδ² * αδ = αβγδ³

Eighth term: βγδ² * βγ = β²γ²δ²

So altogether, the product S1*S2*S3 is equal to:

α³βγδ + α²β²γ² + α²β²δ² + αβ³γδ + α²γ²δ² + αβγ³δ + αβγδ³ + β²γ²δ²

Now, let me simplify each term using the known values:

First, note that αβγδ = 2. Therefore:

Term1: α³βγδ = α²*(αβγδ) = α²*2

Similarly, Term4: αβ³γδ = β²*(αβγδ) = β²*2

Term6: αβγ³δ = γ²*(αβγδ) = γ²*2

Term7: αβγδ³ = δ²*(αβγδ) = δ²*2

Similarly, the other terms:

Term2: α²β²γ² = (αβγ)^2 / α^something? Wait, but maybe express in terms of known variables.

Alternatively, note that Term2 is (αβ)^2 * γ²? Not sure. Wait Term2 is α²β²γ², which is (αβγ)^2. Similarly Term3 is α²β²δ², which is (αβδ)^2. Term5 is α²γ²δ² = (αγδ)^2. Term8 is β²γ²δ² = (βγδ)^2.

But since we know that αβγδ = 2, then:

For example, Term2: (αβγ)^2 = (αβγ)^2 = ?

But also, we know that αβγ + αβδ + αγδ + βγδ = 0 (from Vieta's). Also, each of these terms is (αβγ), etc.

But αβγ = (αβγδ)/δ = 2/δ. Similarly, αβδ = 2/γ, αγδ = 2/β, βγδ = 2/α. Therefore:

Term2: (αβγ)^2 = (2/δ)^2 = 4 / δ²

Similarly:

Term3: (αβδ)^2 = (2/γ)^2 = 4 / γ²

Term5: (αγδ)^2 = (2/β)^2 = 4 / β²

Term8: (βγδ)^2 = (2/α)^2 = 4 / α²

Therefore, terms 2, 3, 5, 8 can be written as 4/δ², 4/γ², 4/β², 4/α².

So let me rewrite the entire expression:

S1*S2*S3 = [2α² + 2β² + 2γ² + 2δ²] + [α²β²γ² + α²β²δ² + α²γ²δ² + β²γ²δ²]

Wait, no. Wait:

Wait, the first four terms:

Term1 + Term4 + Term6 + Term7 = 2α² + 2β² + 2γ² + 2δ²

And the remaining terms:

Term2 + Term3 + Term5 + Term8 = 4/α² + 4/β² + 4/γ² + 4/δ²

Therefore,

S1*S2*S3 = 2(α² + β² + γ² + δ²) + 4(1/α² + 1/β² + 1/γ² + 1/δ²)

We already know that α² + β² + γ² + δ² = P_2 = 4, so first term is 2*4 = 8.

Now, need to compute 1/α² + 1/β² + 1/γ² + 1/δ².

Note that 1/α² + 1/β² + 1/γ² + 1/δ² = (1/α + 1/β + 1/γ + 1/δ)^2 - 2(1/(αβ) + 1/(αγ) + 1/(αδ) + 1/(βγ) + 1/(βδ) + 1/(γδ))

Let me compute this.

First, compute sum of 1/α:

Let me denote Σ1/α. Since Σ1/α = (βγδ + αγδ + αβδ + αβγ)/αβγδ.

From Vieta's:

Σ1/α = (αβγ + αβδ + αγδ + βγδ)/αβγδ = S3 / S4 = 0 / 2 = 0.

Therefore, Σ1/α = 0.

Therefore, Σ1/α² = (Σ1/α)^2 - 2Σ(1/(αβ)) = 0 - 2*(Σ1/(αβ)).

So need to compute Σ1/(αβ).

Σ1/(αβ) is the sum over all pairs of 1/(αβ). Which is equal to [Σ(αγδ + αβδ + αβγ)] / (αβγδ) ?

Wait, let me think. For example, 1/(αβ) + 1/(αγ) + 1/(αδ) + 1/(βγ) + 1/(βδ) + 1/(γδ).

Alternatively, note that Σ1/(αβ) = [Σ(γδ)] / (αβγδ). Because each term 1/(αβ) = γδ / (αβγδ). Therefore:

Σ1/(αβ) = [γδ + αδ + αγ + βδ + βγ + αβ] / (αβγδ). Wait, but the numerator is Σγδ = sum of products two at a time, which is 0.

Wait, but Σγδ = 0. Therefore Σ1/(αβ) = 0 / 2 = 0.

Wait, so Σ1/(αβ) = 0. Therefore Σ1/α² = 0 - 2*0 = 0? That can't be right because if Σ1/α² = 0 then 4*0 = 0. But that would make S1*S2*S3 = 8 + 4*0 = 8? But then R = 8? Wait, but that seems possible. Wait, let me check this again.

Wait, Σ1/α² = (Σ1/α)^2 - 2Σ1/(αβ). We had Σ1/α = 0 and Σ1/(αβ) = [Σγδ]/(αβγδ) = 0/2 = 0. Therefore Σ1/α² = 0 - 2*0 = 0. So that would make the second part of S1*S2*S3 equal to 4*0 = 0. Therefore, total S1*S2*S3 = 8 + 0 = 8. Therefore R = 8. Therefore, the cubic equation is x³ - 8x - 8 = 0.

Wait, but let me confirm this calculation again step by step, because that seems surprising.

First, Σ1/α² = (Σ1/α)^2 - 2Σ1/(αβ).

Σ1/α is 0, as we calculated.

Σ1/(αβ) = sum over all pairs of 1/(αβ). Let me compute that.

Note that each term 1/(αβ) = γδ / (αβγδ). Since αβγδ = 2, so each term 1/(αβ) = γδ / 2. Therefore Σ1/(αβ) = (Σγδ)/2. But Σγδ is sum of products two at a time, which is 0. Hence Σ1/(αβ) = 0 / 2 = 0. Therefore yes, Σ1/α² = 0 - 2*0 = 0.

Hence, S1*S2*S3 = 2*(α² + β² + γ² + δ²) + 4*(Σ1/α²) = 2*4 + 4*0 = 8 + 0 = 8. Therefore R = 8.

So the cubic equation for S1, S2, S3 is x³ - 8x - 8 = 0.

Wait, but let me check if this is correct. If we have S1 + S2 + S3 = 0, S1*S2 + S1*S3 + S2*S3 = -8, and S1*S2*S3 = 8. Then the cubic equation would be x³ - (S1 + S2 + S3)x² + (S1S2 + S1S3 + S2S3)x - S1S2S3 = x³ - 0x² + (-8)x - 8 = x³ - 8x - 8 = 0.

So, solving x³ - 8x - 8 = 0.

Now, need to find the roots of this cubic equation.

Let me try rational roots. Possible rational roots are ±1, ±2, ±4, ±8.

Testing x=2: 8 - 16 -8 = -16 ≠ 0

x=-2: -8 + 16 -8 = 0. Oh! x=-2 is a root.

Therefore, we can factor x³ - 8x - 8 as (x + 2)(x² - 2x - 4).

Let me check:

(x + 2)(x² - 2x - 4) = x³ - 2x² - 4x + 2x² - 4x - 8 = x³ - 8x - 8. Yes, correct.

Therefore, the roots are x = -2 and roots of x² - 2x - 4 = 0. Solving x² - 2x - 4 = 0:

x = [2 ± sqrt(4 + 16)] / 2 = [2 ± sqrt(20)] / 2 = [2 ± 2sqrt(5)] / 2 = 1 ± sqrt(5).

Therefore, the three roots are -2, 1 + sqrt(5), and 1 - sqrt(5). Therefore, the unordered set {S1, S2, S3} is { -2, 1 + sqrt(5), 1 - sqrt(5) }

But wait, hold on. Let me check if that's correct. Since we had S1 + S2 + S3 = 0, and the roots are -2, 1 + sqrt(5), 1 - sqrt(5). Let me check the sum:

-2 + (1 + sqrt(5)) + (1 - sqrt(5)) = -2 + 1 + 1 + sqrt(5) - sqrt(5) = 0. Correct. And the product S1*S2*S3 = (-2)(1 + sqrt(5))(1 - sqrt(5)) = (-2)(1 - 5) = (-2)(-4) = 8. Which matches our earlier calculation. Also, the sum S1*S2 + S1*S3 + S2*S3 was -8. Let me check if with the roots, we have:

For roots a, b, c: a + b + c = 0, ab + ac + bc = -8, abc = 8.

Let me check with a=-2, b=1 + sqrt(5), c=1 - sqrt(5):

ab + ac + bc = (-2)(1 + sqrt(5)) + (-2)(1 - sqrt(5)) + (1 + sqrt(5))(1 - sqrt(5))

= -2 - 2sqrt(5) -2 + 2sqrt(5) + (1 - 5)

= (-4) + (-4) = -8. Correct.

Therefore, yes, the roots are indeed -2, 1 + sqrt(5), 1 - sqrt(5).

Therefore, the required unordered set is { -2, 1 + sqrt(5), 1 - sqrt(5) }.

But let me check if there is any mistake here. Because the question says that α₁, α₂, α₃, α₄ are distinct roots. But does that affect anything? Since the cubic equation has three distinct roots (since discriminant of x² - 2x - 4 is positive), so yes. Therefore, the set { -2, 1 + sqrt(5), 1 - sqrt(5) } is the answer.

But wait, let me check if there is another way to get the answer. For example, maybe I made a mistake in calculation of S1*S2*S3. Let me cross-verify.

Earlier, we had S1*S2*S3 = 8. But also, if S1, S2, S3 are the roots of x³ - 8x - 8 = 0, then yes. But maybe I can think of another approach.

Alternatively, perhaps using the fact that each S_i is of the form p + q where pq = 2. Therefore, each S_i satisfies t² - S_i t + 2 = 0. So for each S_i, S_i² - 2 = S_i * q, where q = 2 / p. Wait, not sure.

Alternatively, since each S_i = p + q, with pq = 2, then p and q are roots of x² - S_i x + 2 = 0. Therefore, for each S_i, the discriminant is S_i² - 8. Therefore, if the roots are real or complex. But since we have three S_i values, which are -2, 1 + sqrt(5), 1 - sqrt(5), then discriminant for S_i=-2 is (-2)^2 - 8 = 4 -8 = -4 < 0. So the corresponding p and q are complex conjugates. For S_i = 1 ± sqrt(5), discriminant is (1 ± sqrt(5))² - 8 = 1 ± 2 sqrt(5) + 5 - 8 = (6 ± 2 sqrt(5)) - 8 = -2 ± 2 sqrt(5).

Wait, for S_i = 1 + sqrt(5):

Discriminant is (1 + sqrt(5))² - 8 = 1 + 2 sqrt(5) + 5 - 8 = (6 + 2 sqrt(5)) - 8 = -2 + 2 sqrt(5) ≈ -2 + 4.47 ≈ 2.47 > 0. Therefore, real roots. Similarly for S_i = 1 - sqrt(5), discriminant is (1 - sqrt(5))² - 8 = 1 - 2 sqrt(5) + 5 -8 = (6 - 2 sqrt(5)) - 8 = -2 - 2 sqrt(5) < 0. Therefore, complex roots.

But since our original equation x⁴ + 2x³ + 2 = 0 has four roots, some of which could be real or complex. However, the question says they are distinct roots. However, regardless of their nature, the set {S_i} is as we found. Therefore, the answer is { -2, 1 + sqrt(5), 1 - sqrt(5) }.

But let me check another way. Let me assume that one of the S_i is -2. For example, suppose that S1 = αβ + γδ = -2. Let me check if that is possible.

If αβ + γδ = -2, and since αβ * γδ = 2, then we have that αβ and γδ are roots of the equation t² - (-2)t + 2 = 0 ⇒ t² + 2t + 2 = 0. The roots are t = [-2 ± sqrt(4 - 8)] / 2 = -1 ± i. Therefore, αβ and γδ are complex numbers. Similarly, for the other S_i's.

But if two of the S_i are 1 ± sqrt(5), which are real numbers, then that would mean that the pair products αβ, γδ are real numbers. Wait, but αβ is a product of two roots. If the equation has four roots, some of which could be complex.

But maybe there are two real roots and two complex roots? Or all complex roots? Let me check the original equation.

Original equation is x⁴ + 2x³ + 2 = 0. Let me check for real roots. Let me consider f(x) = x⁴ + 2x³ + 2. For x real:

- When x approaches ±∞, x⁴ dominates, so f(x) approaches +infty.

At x=0: f(0) = 0 + 0 + 2 = 2 > 0.

At x = -1: f(-1) = 1 - 2 + 2 = 1 > 0.

At x = -2: f(-2) = 16 - 16 + 2 = 2 > 0.

So f(x) is always positive? Wait, but that would imply that there are no real roots, which would mean all roots are complex. Therefore, since the equation is degree 4 with real coefficients, complex roots come in conjugate pairs. Therefore, there are two pairs of complex conjugate roots.

Therefore, the products αβ and γδ would be complex numbers, but their sum S1 = αβ + γδ. However, if αβ and γδ are conjugate pairs, then S1 would be real. Similarly for S2 and S3.

Wait, but according to our previous result, one of the S_i is -2, which is real, and others are 1 ± sqrt(5), which are also real. Therefore, all three S_i are real, which is possible because they are sums of complex conjugates.

But let me check if the equation x³ - 8x - 8 = 0 has three real roots? Well, we have one real root -2 and two other roots 1 ± sqrt(5). Wait, sqrt(5) is approx 2.236, so 1 + sqrt(5) ≈ 3.236, and 1 - sqrt(5) ≈ -1.236. Therefore, all three roots are real. Therefore, the set is { -2, 1 + sqrt(5), 1 - sqrt(5) }, which are three real numbers.

But if the original equation has four complex roots, how can the expressions S_i be real? Because they are sums of products of pairs of complex conjugate roots. For example, if α and β are complex conjugates, and γ and δ are complex conjugates, then αβ and γδ are real numbers, so their sum S1 is real. Similarly for S2 and S3. However, how do S2 and S3 end up being real?

Let me think. Suppose the roots are α, α*, γ, γ*, where * denotes complex conjugate.

Then S1 = αα* + γγ* = |α|² + |γ|², which is real.

S2 = αγ + α*γ*.

Similarly, S3 = αγ* + α*γ.

But S2 and S3 are complex numbers unless αγ is real. Wait, unless αγ is real or something else. Wait, if αγ is real, then S2 and S3 would be real? Let me think.

Suppose α and γ are complex numbers such that α = a + ib, γ = c + id. Then αγ = (a + ib)(c + id) = (ac - bd) + i(ad + bc). For αγ to be real, ad + bc = 0. Similarly, if α and γ are such that their product is real.

Alternatively, maybe not. However, regardless, S2 and S3 may not necessarily be real. However, in our case, we have derived that S2 and S3 are real. Therefore, this suggests that the expressions are real.

But maybe it's not necessary to check further. Given that we followed the algebra correctly and verified that the cubic equation has roots as derived, and that the symmetric sums check out, then the answer is { -2, 1 + sqrt(5), 1 - sqrt(5) }.

But let me check with an alternative approach. Suppose that the equation can be factored.

Original equation: x⁴ + 2x³ + 2 = 0.

Maybe try factoring into quadratics.

Assume that x⁴ + 2x³ + 2 = (x² + a x + b)(x² + c x + d)

Multiply out: x^4 + (a + c)x³ + (ac + b + d)x² + (ad + bc)x + b d

Set equal to x⁴ + 2x³ + 0x² + 0x + 2. Therefore:

1. a + c = 2

2. ac + b + d = 0

3. ad + bc = 0

4. b d = 2

So need to solve for a, b, c, d.

From 1: c = 2 - a

From 4: b d = 2. Let me assume that b and d are real numbers. Since the equation has no real roots (as we saw earlier), then b and d are positive or negative?

If b d = 2, they have the same sign. Let me suppose they are both positive. Let me attempt to solve.

From 3: a d + b c = 0. Substitute c = 2 - a:

a d + b (2 - a) = 0 ⇒ a d - 2b + a b = 0 ⇒ a (d + b) = 2b ⇒ a = (2b)/(d + b)

From equation 2: a c + b + d = 0. Substitute c = 2 - a:

a (2 - a) + b + d = 0.

Let me express this in terms of b and d. Since a = 2b/(d + b), let me substitute:

[2b/(d + b)]*(2 - 2b/(d + b)) + b + d = 0.

This looks complicated, but maybe assume some values for b and d. Since b d = 2. Let me suppose that b = d. Then b² = 2 ⇒ b = √2. So, b = d = √2. Then:

From equation 3: a d + b c = 0 ⇒ a * √2 + √2 * c = 0 ⇒ a + c = 0. But from equation 1: a + c = 2 ⇒ 0 = 2? Contradiction. Not possible.

Alternatively, assume that b = 1, d = 2. Then check equations:

From equation 3: a*2 + 1*c = 0 ⇒ 2a + c = 0. But from equation 1, c = 2 - a. Therefore:

2a + 2 - a = 0 ⇒ a + 2 = 0 ⇒ a = -2. Then c = 2 - (-2) = 4.

Check equation 2: a c + b + d = (-2)(4) + 1 + 2 = -8 + 3 = -5 ≠ 0. Not good.

Alternatively, try b = 2, d = 1:

From equation 3: a*1 + 2*c = 0 ⇒ a + 2c = 0. From equation 1: c = 2 - a. Substitute:

a + 2*(2 - a) = 0 ⇒ a + 4 - 2a = 0 ⇒ -a + 4 = 0 ⇒ a = 4. Then c = 2 - 4 = -2.

Check equation 2: a c + b + d = 4*(-2) + 2 + 1 = -8 + 3 = -5 ≠ 0. Again no good.

Alternatively, try b= -1, d=-2. Then b d = (-1)(-2) = 2.

Equation 3: a*(-2) + (-1)*c = 0 ⇒ -2a - c = 0 ⇒ c = -2a.

From equation 1: a + c = 2 ⇒ a - 2a = 2 ⇒ -a = 2 ⇒ a = -2 ⇒ c = -2*(-2) = 4.

Check equation 2: a c + b + d = (-2)(4) + (-1) + (-2) = -8 - 3 = -11 ≠ 0. Not good.

Alternatively, maybe b and d are not rational? Let me consider solving equations:

From above, b d = 2, and let me express a and c in terms of b and d.

From equation 3: a d + b c = 0, and c = 2 - a. Therefore, a d + b(2 - a) = 0 ⇒ a(d - b) + 2b = 0 ⇒ a = -2b / (d - b)

From equation 2: a(2 - a) + b + d = 0. Substitute a:

Let me denote a = -2b / (d - b). Let me denote that d = 2 / b (from b d = 2). Therefore:

a = -2b / ( (2 / b) - b ) = -2b / ( (2 - b²)/b ) = -2b * (b / (2 - b²)) ) = -2b² / (2 - b²)

Now substitute into equation 2:

a(2 - a) + b + d = 0

First compute 2 - a = 2 - [ -2b² / (2 - b²) ] = 2 + 2b² / (2 - b²) = [ 2(2 - b²) + 2b² ] / (2 - b²) ) = [4 - 2b² + 2b² ] / (2 - b²) ) = 4 / (2 - b²)

Now compute a(2 - a) = [ -2b² / (2 - b²) ] * [ 4 / (2 - b²) ] = -8b² / (2 - b²)^2

Then equation 2: -8b² / (2 - b²)^2 + b + d = 0. Since d = 2 / b:

=> -8b² / (2 - b²)^2 + b + 2 / b = 0

Multiply both sides by b(2 - b²)^2 to eliminate denominators:

-8b³ + b(2 - b²)^2 * b + 2(2 - b²)^2 = 0

Wait, this seems complicated. Let me compute term by term:

Let me denote term1 = -8b² / (2 - b²)^2 * b(2 - b²)^2 = -8b³

term2 = [b + 2/b] * b(2 - b²)^2 = b^2(2 - b²)^2 + 2(2 - b²)^2

Therefore, equation becomes:

-8b³ + b²(2 - b²)^2 + 2(2 - b²)^2 = 0

Factor out (2 - b²)^2:

[ b² + 2 ] (2 - b²)^2 - 8b³ = 0

Let me expand (2 - b²)^2 = 4 - 4b² + b⁴

So:

[ b² + 2 ] (4 - 4b² + b⁴) - 8b³ = 0

Let me compute this:

First, expand [ b² + 2 ] * [ b⁴ - 4b² + 4 ]:

= b²*(b⁴ - 4b² + 4) + 2*(b⁴ - 4b² + 4)

= b^6 - 4b^4 + 4b² + 2b^4 - 8b² + 8

= b^6 - 2b^4 - 4b² + 8

Then subtract 8b³:

Total expression: b^6 - 2b^4 - 4b² + 8 - 8b³ = b^6 - 2b^4 - 8b³ - 4b² + 8 = 0

This is a sixth-degree equation in b, which is extremely complex. This seems not helpful. Therefore, factoring might not be the way to go here.

Therefore, returning to our previous result, since we have derived that the set is { -2, 1 + sqrt(5), 1 - sqrt(5) }, and all the symmetric sums check out, and the cubic equation is correct, I think this is the correct answer.

Thus, the unordered set is { -2, 1 + √5, 1 - √5 }.

But let me confirm once more with another approach. Suppose we take specific values. Let me try to compute one of the S_i.

For example, suppose we can find the value of S1 = αβ + γδ.

From Vieta, we know that αβγδ = 2. Also, we have that S1 = αβ + γδ. Let me denote p = αβ, q = γδ. Then pq = 2, and S1 = p + q. So p and q are roots of t² - S1 t + 2 = 0.

Also, from Vieta's formulas, we know that α + β + γ + δ = -2. Let me denote that α + β = m, γ + δ = n. Then m + n = -2. Also, from Vieta's, we have αβ + γδ = S1, and also αβ + αγ + αδ + βγ + βδ + γδ = 0.

So, αβ + γδ + αγ + αδ + βγ + βδ = 0 ⇒ S1 + [αγ + αδ + βγ + βδ] = 0.

But αγ + αδ + βγ + βδ = α(γ + δ) + β(γ + δ) = (α + β)(γ + δ) = m n. So:

S1 + m n = 0.

But since m + n = -2, and m n = ?

Also, note that m = α + β, and n = γ + δ.

Also, the products:

αβ = p, γδ = q.

Additionally, the sum of α + β + γ + δ = m + n = -2.

Also, for the quadratic equations:

For α and β: x² - m x + p = 0

For γ and δ: x² - n x + q = 0

Since the original equation is x⁴ + 2x³ + 2 = 0. Therefore, the quartic can be written as (x² - m x + p)(x² - n x + q) = 0. Expanding this gives:

x⁴ - (m + n)x³ + (mn + p + q)x² - (mq + np)x + pq = 0.

Compare to original equation x⁴ + 2x³ + 0x² + 0x + 2 = 0. Therefore:

- (m + n) = 2 ⇒ m + n = -2 (which matches)

mn + p + q = 0 ⇒ mn + S1 = 0 (since p + q = S1)

- (mq + np) = 0 ⇒ mq + np = 0

pq = 2

From before, we have mn + S1 = 0. Since m + n = -2, we can express mn in terms of m or n. Let me denote mn + S1 = 0 ⇒ mn = -S1. Also, since m + n = -2, we can write mn = -S1.

Additionally, mq + np = 0. But since p = αβ, q = γδ. Also, q = 2/p from pq = 2. So substitute:

mq + np = m*(2/p) + n*p = 0 ⇒ (2m)/p + n p = 0.

But we can express n = -2 - m, so:

(2m)/p + (-2 - m)p = 0 ⇒ (2m)/p - 2p - m p = 0 ⇒ Multiply both sides by p:

2m - 2p² - m p² = 0 ⇒ 2m - 2p² - m p² = 0.

Also, from mn = -S1 and m + n = -2, so mn = -S1 ⇒ m(-2 - m) = -S1 ⇒ -2m - m² = -S1 ⇒ S1 = m² + 2m.

Therefore, S1 = m² + 2m.

But also, from the equation above: 2m - 2p² - m p² = 0.

But since p = αβ, and we know that αβ is a root of t² - m t + p = 0. Wait, but p is also related to m. Wait, this seems too convoluted.

Alternatively, since S1 = m² + 2m and also S1 = p + q = p + 2/p. Let me write:

S1 = p + 2/p = m² + 2m.

So we have p + 2/p = m² + 2m. Also, since p is a root of the quadratic x² - m x + p = 0, so p² - m p + p = 0? Wait, no. Wait, the quadratic equation is x² - m x + p = 0, so the roots are α and β. But p = αβ. So from quadratic equation, the product of roots is p, which is given. Also, sum of roots is m.

But I don't see a direct relation between p and m. However, we have that p is also related through the equation above.

But this seems too tangled. Maybe pick an S1 value and check consistency. For example, suppose S1 = -2. Then S1 = m² + 2m = -2 ⇒ m² + 2m + 2 = 0 ⇒ m = [-2 ± sqrt(4 - 8)]/2 = -1 ± i. Therefore, m is complex. Which makes sense, since if the roots are complex, then m = α + β is a complex number.

Then, p + 2/p = -2. Let me solve for p:

p + 2/p = -2 ⇒ p² + 2 = -2p ⇒ p² + 2p + 2 = 0 ⇒ p = [-2 ± sqrt(4 - 8)] / 2 = -1 ± i. Which matches with the roots of the quadratic x² - m x + p = 0. Indeed, since m is -1 + i or -1 - i, then the quadratic would be x² - m x + p = 0, with m and p both complex.

This seems consistent.

Similarly, if we take S1 = 1 + sqrt(5), then S1 = m² + 2m. Let me solve:

m² + 2m - S1 = 0 ⇒ m = [-2 ± sqrt(4 + 4S1)] / 2 = [-2 ± 2sqrt(1 + S1)] / 2 = -1 ± sqrt(1 + S1).

For S1 = 1 + sqrt(5), this gives m = -1 ± sqrt(1 + 1 + sqrt(5)) = -1 ± sqrt(2 + sqrt(5)). Similarly, for S1 = 1 - sqrt(5), m = -1 ± sqrt(1 + 1 - sqrt(5)) = -1 ± sqrt(2 - sqrt(5)).

But since our original equation has no real roots, these m would be complex or what? Wait, sqrt(2 + sqrt(5)) is real, because sqrt(5) ≈ 2.236, so 2 + sqrt(5) ≈ 4.236, so sqrt of that is ≈ 2.06, real. Similarly, sqrt(2 - sqrt(5)) is sqrt of 2 - 2.236 ≈ -0.236, which is imaginary. Therefore, for S1 = 1 - sqrt(5), m would involve sqrt of negative number, meaning m is complex. Which again is consistent.

Therefore, all these S_i values are consistent. Therefore, the answer is indeed { -2, 1 + sqrt(5), 1 - sqrt(5) }.

Thus, the unordered set is { -2, 1 + √5, 1 - √5 }.

**Final Answer**
\boxed{\{-2, 1 - \sqrt{5}, 1 + \sqrt{5}\}}

To determine the unordered set
$$
\{\alpha_1\alpha_2 + \alpha_3\alpha_4, \alpha_1\alpha_3 + \alpha_2\alpha_4, \alpha_1\alpha_4 + \alpha_2\alpha_3\}
$$
where $ \alpha_1, \alpha_2, \alpha_3, \alpha_4 $ are the four **distinct** roots of the polynomial equation
$$
x^4 + 2x^3 + 0x^2 + 0x + 2 = 0,
$$
we employ **Vieta's formulas** to simplify the situation.

---

### 1. **Apply Vieta's Formulas**

Let the roots be $ \alpha_1, \alpha_2, \alpha_3, \alpha_4 $. The quartic equation can be written as
$$
x^4 + 2x^3 + 2 = 0.
$$
This corresponds to standard coefficients:
- $ a = 2 $
- $ b = 0 $
- $ c = 0 $
- $ d = 2 $

From Vieta’s formulas for $ x^4 + a x^3 + b x^2 + c x + d = 0 $, we have:

- $ \alpha_1 + \alpha_2 + \alpha_3 + \alpha_4 = -a = -2 $
- $ \alpha_1\alpha_2 + \alpha_1\alpha_3 + \alpha_1\alpha_4 + \alpha_2\alpha_3 + \alpha_2\alpha_4 + \alpha_3\alpha_4 = b = 0 $
- $ \alpha_1\alpha_2\alpha_3 + \alpha_1\alpha_2\alpha_4 + \alpha_1\alpha_3\alpha_4 + \alpha_2\alpha_3\alpha_4 = -c = 0 $
- $ \alpha_1\alpha_2\alpha_3\alpha_4 = d = 2 $

---

### 2. **Target Expression**

We define:
$$
S_1 = \alpha_1\alpha_2 + \alpha_3\alpha_4, \quad
S_2 = \alpha_1\alpha_3 + \alpha_2\alpha_4, \quad
S_3 = \alpha_1\alpha_4 + \alpha_2\alpha_3
$$

Our goal is to compute the set $ \{S_1, S_2, S_3\} $.

---

### 3. **Symmetric Sum Relationships**

It's known that the sum of all two-product pairs of roots is $ \sum_{1 \leq i < j \leq 4} \alpha_i\alpha_j = 0 $.
Thus,
$$
S_1 + S_2 + S_3 = 0.
$$
So the sum of these three values is zero.

Let us denote:

- $ S = S_1 + S_2 + S_3 = 0 $
- $ Q = S_1S_2 + S_1S_3 + S_2S_3 $
- $ R = S_1S_2S_3 $

Then, the three values $ S_1, S_2, S_3 $ are roots of the cubic equation:
$$
x^3 - Sx^2 + Qx - R = x^3 - 8x - 8 = 0
$$

This cubic equation is found by using Newton’s identities after computing the power sums of the original quartic roots.

---

### 4. **Solve the Cubic Equation**

We solve:
$$
x^3 - 8x - 8 = 0
$$

Try $ x = -2 $:
$$
(-2)^3 - 8(-2) - 8 = -8 + 16 - 8 = 0
$$
So $ x = -2 $ is a root.

Factor:
$$
x^3 - 8x - 8 = (x + 2)(x^2 - 2x - 4)
$$

Solve the quadratic:
$$
x^2 - 2x - 4 = 0 \Rightarrow x = \frac{2 \pm \sqrt{4 + 16}}{2} = \frac{2 \pm \sqrt{20}}{2} = 1 \pm \sqrt{5}
$$

So the three roots of the cubic are:
$$
x = -2, \quad 1 + \sqrt{5}, \quad 1 - \sqrt{5}
$$

---

### 5. **Conclusion**

The unordered set $ \{S_1, S_2, S_3\} $ consists of these three values:

$$
\boxed{ \{-2, 1 - \sqrt{5}, 1 + \sqrt{5} \} }
$$

[Assistant]
Explanation:
Okay, so I need to find the unordered set {α₁α₂ + α₃α₄, α₁α₃ + α₂α₄, α₁α₄ + α₂α₃} where α₁, α₂, α₃, α₄ are the roots of the equation x⁴ + 2x³ + 2 = 0. Hmm, let me recall Vieta's formulas for quartic equations.

First, for a quartic equation x⁴ + a x³ + b x² + c x + d = 0, the sums of roots are related to the coefficients. Specifically, if the roots are α, β, γ, δ, then:

α + β + γ + δ = -a

αβ + αγ + αδ + βγ + βδ + γδ = b

αβγ + αβδ + αγδ + βγδ = -c

αβγδ = d

In my given equation, the equation is x⁴ + 2x³ + 0x² + 0x + 2 = 0. Wait, so the coefficients are:

a = 2, b = 0, c = 0, d = 2.

So applying Vieta's formulas:

Sum of roots: α₁ + α₂ + α₃ + α₄ = -a = -2

Sum of products two at a time: α₁α₂ + α₁α₃ + α₁α₄ + α₂α₃ + α₂α₄ + α₃α₄ = b = 0

Sum of products three at a time: α₁α₂α₃ + α₁α₂α₄ + α₁α₃α₄ + α₂α₃α₄ = -c = 0

Product of roots: α₁α₂α₃α₄ = d = 2

Okay, so we need to compute the three expressions:

S1 = α₁α₂ + α₃α₄

S2 = α₁α₃ + α₂α₄

S3 = α₁α₄ + α₂α₃

And then put them in an unordered set {S1, S2, S3}

So, the question is to find these three values.

Now, note that S1 + S2 + S3 = (α₁α₂ + α₃α₄) + (α₁α₃ + α₂α₄) + (α₁α₄ + α₂α₃) = sum of all products two at a time, which is given by Vieta as 0. Therefore, S1 + S2 + S3 = 0.

So the three numbers add up to zero. But we need to find each of them.

But how can we find individual values?

Alternatively, maybe we can think of them as the sums of products of pairs of roots, grouped into two pairs.

Let me think. Suppose we consider that the roots are grouped into pairs. For example, S1 is the sum of products of two pairs. However, since there are three such expressions, each corresponding to a different way of grouping the roots into two pairs.

But perhaps I can relate these expressions to symmetric sums?

Alternatively, maybe I can consider variables. Let me denote:

Let me think of the roots as α, β, γ, δ. Let me denote the three expressions as:

S1 = αβ + γδ

S2 = αγ + βδ

S3 = αδ + βγ

So, if we can find S1, S2, S3, then we can put them in the set.

But how?

Since we know that S1 + S2 + S3 = 0. Also, maybe we can compute S1*S2 + S1*S3 + S2*S3? Or perhaps other combinations? Alternatively, maybe we can find another equation involving S1, S2, S3.

Alternatively, perhaps consider that if we can find expressions for S1, S2, S3 in terms of the known Vieta's formulas.

Alternatively, perhaps consider that if we take the product (x - α)(x - β)(x - γ)(x - δ) = x⁴ + 2x³ + 2.

But maybe another approach: Let me think about the roots. Let me suppose that the roots come in pairs? Not sure.

Alternatively, perhaps consider that if we can find S1, S2, S3, maybe we can relate them through some equations. Let me try:

Let me consider that:

Let me denote S = S1 + S2 + S3 = 0.

Let me denote P = product of roots = αβγδ = 2.

Also, the sum of products two at a time is zero. Let me see:

Also, let me think about the sum S1*S2 + S1*S3 + S2*S3?

But that seems complicated. Alternatively, maybe think about individual S1, S2, S3. Let me try squaring the sum of roots, for example.

We know that (α + β + γ + δ)² = α² + β² + γ² + δ² + 2(αβ + αγ + αδ + βγ + βδ + γδ)

But we know that α + β + γ + δ = -2, so (-2)² = 4 = (α² + β² + γ² + δ²) + 2*(0) => α² + β² + γ² + δ² = 4.

But not sure if that helps yet.

Alternatively, maybe think about the values S1, S2, S3. Let me take S1 = αβ + γδ. Maybe if I can find expressions for αβ, γδ, etc.?

Alternatively, let me think of S1, S2, S3 as the sums of two different products. Let me think about variables. Let me suppose that:

Let me denote P1 = αβ, P2 = γδ. Then S1 = P1 + P2.

Similarly, for S2 and S3, but they would involve different pairings.

Alternatively, perhaps consider that if I can find P1, P2, etc. But maybe not.

Alternatively, think about the equations involving Vieta's formulas. Let me think. Let me also note that:

From Vieta's formula, we know that:

αβ + αγ + αδ + βγ + βδ + γδ = 0.

But also, if we take S1 = αβ + γδ, then S2 = αγ + βδ, and S3 = αδ + βγ.

So, if we add S1 + S2 + S3 = 0 as before. Now, if I can find other relations between S1, S2, S3.

Let me also note that:

Suppose we take S1 = αβ + γδ. If I multiply S1 by something?

Alternatively, consider that maybe S1 * something?

Alternatively, think about the product of the roots. Since αβγδ = 2, so (αβ)(γδ) = 2. Therefore, if S1 = αβ + γδ, then S1 is the sum of two numbers whose product is 2. Similarly for S2 and S3.

Wait, but S2 = αγ + βδ. Let me check if (αγ)(βδ) = αβγδ = 2. Similarly for S3: (αδ)(βγ) = αβγδ = 2. So each of the S1, S2, S3 is the sum of two numbers whose product is 2. Interesting!

Therefore, each S1, S2, S3 satisfies the equation t² - (sum)t + 2 = 0. Wait, but each S_i is a sum of two numbers whose product is 2. So, for example, S1 = p + q, where pq = 2. Therefore, p and q are roots of the equation x² - S1 x + 2 = 0.

But maybe this is not directly helpful? Hmm.

But perhaps if we can find S1, S2, S3. Let me consider that S1, S2, S3 are roots of some cubic equation? Maybe?

Alternatively, since we have S1 + S2 + S3 = 0, and also perhaps S1S2 + S1S3 + S2S3 and S1S2S3?

If I can find these two other symmetric sums, then I can find the set {S1, S2, S3}.

So let me try to compute S1S2 + S1S3 + S2S3 and S1S2S3.

First, let me compute S1S2 + S1S3 + S2S3.

Alternatively, note that (S1 + S2 + S3)^2 = S1² + S2² + S3² + 2(S1S2 + S1S3 + S2S3). Since S1 + S2 + S3 = 0, this implies that 0 = S1² + S2² + S3² + 2(S1S2 + S1S3 + S2S3), so S1² + S2² + S3² = -2(S1S2 + S1S3 + S2S3). But not sure if helpful yet.

Alternatively, perhaps compute S1S2 + S1S3 + S2S3 directly. Let me see.

Let me expand S1S2:

S1*S2 = (αβ + γδ)(αγ + βδ)

Let me compute this:

First, expand the product:

= αβ*αγ + αβ*βδ + γδ*αγ + γδ*βδ

= α²βγ + αβ²δ + αγ²δ + βγδ²

Similarly, S1*S3 = (αβ + γδ)(αδ + βγ)

= αβ*αδ + αβ*βγ + γδ*αδ + γδ*βγ

= α²βδ + αβ²γ + αδ²γ + βγ²δ

Similarly, S2*S3 = (αγ + βδ)(αδ + βγ)

= αγ*αδ + αγ*βγ + βδ*αδ + βδ*βγ

= α²γδ + αβγ² + αβδ² + β²γδ

Therefore, adding S1*S2 + S1*S3 + S2*S3:

= [α²βγ + αβ²δ + αγ²δ + βγδ²] + [α²βδ + αβ²γ + αδ²γ + βγ²δ] + [α²γδ + αβγ² + αβδ² + β²γδ]

Hmm, this seems complicated, but maybe we can combine like terms.

Let me list all terms:

From S1*S2:

Term1: α²βγ

Term2: αβ²δ

Term3: αγ²δ

Term4: βγδ²

From S1*S3:

Term5: α²βδ

Term6: αβ²γ

Term7: αδ²γ

Term8: βγ²δ

From S2*S3:

Term9: α²γδ

Term10: αβγ²

Term11: αβδ²

Term12: β²γδ

Now, let me see if these terms can be grouped together.

For example, Term1: α²βγ and Term6: αβ²γ. These have αβγ as a common factor.

Similarly, Term2: αβ²δ and Term5: α²βδ have αβδ as a common factor.

Similarly, Term3: αγ²δ and Term7: αδ²γ have αγδ as common factor.

Term4: βγδ² and Term12: β²γδ have βγδ as common factor.

Term8: βγ²δ and Term10: αβγ² have βγ² as common factor?

Wait, Term8 is βγ²δ and Term10 is αβγ². So they share βγ². Similarly Term11: αβδ² and Term4? Not sure.

Alternatively, maybe factor variables.

Alternatively, maybe factor out common terms. Let me see:

Let me consider each term:

Term1: α²βγ

Term2: αβ²δ

Term3: αγ²δ

Term4: βγδ²

Term5: α²βδ

Term6: αβ²γ

Term7: αγδ²

Term8: βγ²δ

Term9: α²γδ

Term10: αβγ²

Term11: αβδ²

Term12: β²γδ

So, maybe group terms with similar variables.

For example, terms involving α²βγ, α²βδ, α²γδ, and similar.

Let me group terms with α²:

- α²βγ (Term1)

- α²βδ (Term5)

- α²γδ (Term9)

Similarly for β²:

- αβ²δ (Term2)

- αβ²γ (Term6)

- β²γδ (Term12)

For γ²:

- αγ²δ (Term3)

- βγ²δ (Term8)

- αβγ² (Term10)

For δ²:

- βγδ² (Term4)

- αγδ² (Term7)

- αβδ² (Term11)

Wait, so each of these groups has three terms, and each group corresponds to variables squared. So, for example, the α² terms are α²βγ, α²βδ, α²γδ. Similarly for others.

So, let me compute the sum:

Sum = [α²βγ + α²βδ + α²γδ] + [αβ²δ + αβ²γ + β²γδ] + [αγ²δ + βγ²δ + αβγ²] + [βγδ² + αγδ² + αβδ²]

Wait, but actually, each variable squared has three terms. However, in the previous list, each of the four terms (α², β², γ², δ²) have three terms? Wait, no. Wait, looking at the above:

- The α² terms: Term1, Term5, Term9

- The β² terms: Term2, Term6, Term12

- The γ² terms: Term3, Term8, Term10

- The δ² terms: Term4, Term7, Term11

So yes, each squared variable has three terms. Therefore, the entire sum is:

Sum = (α²βγ + α²βδ + α²γδ) + (αβ²δ + αβ²γ + β²γδ) + (αγ²δ + βγ²δ + αβγ²) + (βγδ² + αγδ² + αβδ²)

But note that each of these groups is the same as:

For example, the first group: α²βγ + α²βδ + α²γδ = α²β(γ + δ) + α²γδ. Similarly for others. But perhaps there is a better way to express this.

Alternatively, note that each group is similar to the product of α² times the sum of products of β, γ, δ?

Wait, not sure. Alternatively, factor terms.

Alternatively, notice that each term in the sum is of the form (variable squared)*(product of two others). For example, α²βγ is α² times βγ. Similarly for all others.

So, the entire sum can be written as:

Sum = α²βγ + α²βδ + α²γδ + αβ²δ + αβ²γ + β²γδ + αγ²δ + βγ²δ + αβγ² + βγδ² + αγδ² + αβδ²

Which can be grouped as:

= α²(βγ + βδ + γδ) + β²(αδ + αγ + γδ) + γ²(αδ + αβ + βδ) + δ²(αβ + αγ + βγ)

But wait, this seems complicated. Wait, let me check:

For instance, the first term α²(βγ + βδ + γδ). Is that true? Let me see:

α²βγ + α²βδ + α²γδ = α²(βγ + βδ + γδ). Yes. Similarly for the others.

So, the Sum is equal to:

α²(βγ + βδ + γδ) + β²(αδ + αγ + γδ) + γ²(αδ + αβ + βδ) + δ²(αβ + αγ + βγ)

But note that βγ + βδ + γδ is equal to sum of products of β, γ, δ. Let me recall Vieta's formula.

From Vieta's, we know that:

α + β + γ + δ = -2

αβ + αγ + αδ + βγ + βδ + γδ = 0

αβγ + αβδ + αγδ + βγδ = 0

αβγδ = 2

So, if we take βγ + βδ + γδ, this is equal to [αβ + αγ + αδ + βγ + βδ + γδ] - α(β + γ + δ) = 0 - α(β + γ + δ)

Similarly, since α + β + γ + δ = -2, then β + γ + δ = -2 - α. Therefore:

βγ + βδ + γδ = -α(-2 - α) = 2α + α².

Similarly, for the other terms:

For example, αδ + αγ + γδ = -β(-2 - β) = 2β + β²

Similarly, αδ + αβ + βδ = -γ(-2 - γ) = 2γ + γ²

And αβ + αγ + βγ = -δ(-2 - δ) = 2δ + δ²

Therefore, substituting back into the Sum:

Sum = α²(2α + α²) + β²(2β + β²) + γ²(2γ + γ²) + δ²(2δ + δ²)

= 2α³ + α⁴ + 2β³ + β⁴ + 2γ³ + γ⁴ + 2δ³ + δ⁴

= (α⁴ + β⁴ + γ⁴ + δ⁴) + 2(α³ + β³ + γ³ + δ³)

So, we have that S1*S2 + S1*S3 + S2*S3 equals this expression. Therefore, if we can compute α⁴ + β⁴ + γ⁴ + δ⁴ and α³ + β³ + γ³ + δ³, then we can compute this Sum.

Okay, so let's compute α³ + β³ + γ³ + δ³.

There's a formula for the sum of cubes. Recall that:

α³ + β³ + γ³ + δ³ = (α + β + γ + δ)³ - 3(α + β + γ + δ)(αβ + αγ + αδ + βγ + βδ + γδ) + 3(αβγ + αβδ + αγδ + βγδ)

But let me check that formula. For three variables, the formula is a³ + b³ + c³ = (a + b + c)^3 - 3(a + b + c)(ab + bc + ac) + 3abc. For four variables, perhaps there is an analogous formula. Alternatively, maybe use Newton's identities?

Alternatively, since we know the values of the power sums. Let me recall that power sums can be computed using recurrence relations.

Alternatively, since the roots satisfy the equation x⁴ + 2x³ + 2 = 0, so for each root, x⁴ = -2x³ - 2. Therefore, for each root, x⁴ = -2x³ - 2. Therefore, we can use this to express higher powers in terms of lower powers.

Therefore, we can compute α³ + β³ + γ³ + δ³ and α⁴ + β⁴ + γ⁴ + δ⁴.

Let me first compute α³ + β³ + γ³ + δ³.

Let me denote S1 = α + β + γ + δ = -2

S2 = αβ + αγ + αδ + βγ + βδ + γδ = 0

S3 = αβγ + αβδ + αγδ + βγδ = 0

S4 = αβγδ = 2

Let me recall that the sum of cubes can be calculated by:

Σα³ = (Σα)^3 - 3ΣαΣβγ + 3Σαβγ

Wait, for four variables, the formula is:

Σα³ = (Σα)^3 - 3ΣαΣβγ + 3Σαβγ

Wait, actually, let me check:

For n variables, Σα³ can be calculated using the formula:

Σα³ = (Σα)^3 - 3ΣαΣβγ + 3Σαβγ. Hmm, not sure. Let me check with three variables.

For three variables, the formula is a³ + b³ + c³ = (a + b + c)^3 - 3(a + b + c)(ab + bc + ac) + 3abc. So yes, for four variables, perhaps:

Σα³ = (Σα)^3 - 3Σα*(Σβγ) + 3Σαβγ. Let me check if this holds.

Let me consider expanding (Σα)^3:

(Σα)^3 = Σα³ + 3Σα²β + 6Σαβγ

Wait, no. Let me recall that:

(Σα)^3 = Σα³ + 3Σα²β + 6αβγ. Wait, for four variables, it's more complicated. Let me think.

Alternatively, maybe use Newton's identities, which relate power sums to elementary symmetric sums.

Newton's identities relate the power sums P_k = α^k + β^k + γ^k + δ^k with the elementary symmetric polynomials.

For a quartic equation, Newton's identities can be used. Let me recall that.

For k ≥ 1,

P_k + a P_{k-1} + b P_{k-2} + c P_{k-3} + d P_{k-4} = 0

Wait, where the equation is x^4 + a x^3 + b x^2 + c x + d = 0. But in our case, the equation is x⁴ + 2x³ + 0x² + 0x + 2 = 0. Therefore, a=2, b=0, c=0, d=2.

Therefore, Newton's identities state that:

For k ≥ 1,

P_k + 2 P_{k-1} + 0*P_{k-2} + 0*P_{k-3} + 2 P_{k-4} = 0.

But note that for k < 4, P_{k-4} would be zero? Wait, maybe not. Wait, Newton's identities for monic polynomials:

If the polynomial is x^4 + a x^3 + b x^2 + c x + d, then:

P_1 + a = 0

P_2 + a P_1 + 2b = 0

P_3 + a P_2 + b P_1 + 3c = 0

P_4 + a P_3 + b P_2 + c P_1 + 4d = 0

And for k > 4:

P_k + a P_{k-1} + b P_{k-2} + c P_{k-3} + d P_{k-4} = 0

Therefore, let me compute P_1, P_2, P_3, P_4.

Given that:

a = 2, b = 0, c = 0, d = 2.

First, P_1:

From the first identity: P_1 + a = 0 ⇒ P_1 = -a = -2. Which matches our earlier Vieta's formula.

Next, P_2:

Second identity: P_2 + a P_1 + 2b = 0 ⇒ P_2 + 2*(-2) + 0 = 0 ⇒ P_2 - 4 = 0 ⇒ P_2 = 4.

Wait, but earlier I calculated that α² + β² + γ² + δ² = 4? From the square of the sum. Yes, because (α + β + γ + δ)^2 = 4 = sum of squares + 2*(sum of products). But sum of products is 0, so sum of squares = 4. Which matches P_2 = 4. Good.

Next, P_3:

Third identity: P_3 + a P_2 + b P_1 + 3c = 0 ⇒ P_3 + 2*4 + 0*(-2) + 0 = 0 ⇒ P_3 + 8 = 0 ⇒ P_3 = -8.

So the sum of cubes is -8.

Now, P_4:

Fourth identity: P_4 + a P_3 + b P_2 + c P_1 + 4d = 0 ⇒ P_4 + 2*(-8) + 0*4 + 0*(-2) + 4*2 = 0 ⇒ P_4 - 16 + 8 = 0 ⇒ P_4 - 8 = 0 ⇒ P_4 = 8.

Alternatively, since each root satisfies x⁴ = -2x³ -2, then P_4 = Σx⁴ = Σ(-2x³ -2) = -2Σx³ - 2*4 = -2*(-8) - 8 = 16 - 8 = 8. Which matches. Good.

Now, going back to our earlier expression for S1*S2 + S1*S3 + S2*S3 = (α⁴ + β⁴ + γ⁴ + δ⁴) + 2(α³ + β³ + γ³ + δ³) = P_4 + 2P_3 = 8 + 2*(-8) = 8 - 16 = -8.

So S1*S2 + S1*S3 + S2*S3 = -8.

Earlier, we had S1 + S2 + S3 = 0.

Therefore, we have the following for the symmetric sums of S1, S2, S3:

Let me denote S = S1 + S2 + S3 = 0

Q = S1*S2 + S1*S3 + S2*S3 = -8

Now, if I can find R = S1*S2*S3, then I can form the cubic equation whose roots are S1, S2, S3:

x³ - Sx² + Qx - R = 0 ⇒ x³ + 0x² -8x - R = 0 ⇒ x³ -8x - R = 0.

So if we can find R, then we can find the cubic equation, and then find its roots.

So, need to compute R = S1*S2*S3.

How?

Let me think. Let me note that S1 = αβ + γδ, S2 = αγ + βδ, S3 = αδ + βγ.

So, S1*S2*S3 = (αβ + γδ)(αγ + βδ)(αδ + βγ)

This seems complicated, but maybe we can expand it?

Alternatively, perhaps use relations from Vieta's formulas.

Alternatively, note that if we consider the product (αβ + γδ)(αγ + βδ)(αδ + βγ). Maybe there's a clever way to compute this.

Alternatively, think about the product (αβ + γδ)(αγ + βδ)(αδ + βγ). Let me first compute (αβ + γδ)(αγ + βδ).

Earlier, we had expanded this as α²βγ + αβ²δ + αγ²δ + βγδ² + ... Wait, no, earlier expansion for S1*S2 gave terms that led us to P_4 + 2P_3. Wait, no, but here we need to compute S1*S2*S3, which is multiplying three such terms.

Alternatively, perhaps use substitution variables.

Alternatively, think that since αβγδ = 2, so perhaps set variables such that:

Let me consider that for example, let me set:

Let me denote p = αβ, q = γδ, so S1 = p + q. Similarly, for S2 and S3, but need to relate them to p and q.

But S2 = αγ + βδ. Hmm. Let me see:

Let me note that αγ + βδ. If p = αβ, q = γδ, then maybe express S2 in terms of p and q?

Alternatively, perhaps consider that αγ + βδ. Let me think. If we consider product αγ * βδ = (αβ)(γδ) = pq = 2. So similar to S1.

But maybe not helpful yet.

Alternatively, think of αγ + βδ as a combination of terms.

Alternatively, maybe use the fact that α, β, γ, δ are roots of the equation, so we can relate them through symmetric sums.

Alternatively, perhaps note that the product S1*S2*S3 can be expressed in terms of known symmetric sums.

Alternatively, let me try to compute S1*S2*S3.

Let me consider that:

First, compute (αβ + γδ)(αγ + βδ) = αβαγ + αββδ + γδαγ + γδβδ = α²βγ + αβ²δ + αγ²δ + βγδ². Which we had earlier. Then multiply this by (αδ + βγ):

So, multiplying the above by (αδ + βγ):

= [α²βγ + αβ²δ + αγ²δ + βγδ²] * [αδ + βγ]

This is going to be a lot, but let me try.

First, expand term by term:

First term: α²βγ * αδ = α³βγδ

Second term: α²βγ * βγ = α²β²γ²

Third term: αβ²δ * αδ = α²β²δ²

Fourth term: αβ²δ * βγ = αβ³γδ

Fifth term: αγ²δ * αδ = α²γ²δ²

Sixth term: αγ²δ * βγ = αβγ³δ

Seventh term: βγδ² * αδ = αβγδ³

Eighth term: βγδ² * βγ = β²γ²δ²

So altogether, the product S1*S2*S3 is equal to:

α³βγδ + α²β²γ² + α²β²δ² + αβ³γδ + α²γ²δ² + αβγ³δ + αβγδ³ + β²γ²δ²

Now, let me simplify each term using the known values:

First, note that αβγδ = 2. Therefore:

Term1: α³βγδ = α²*(αβγδ) = α²*2

Similarly, Term4: αβ³γδ = β²*(αβγδ) = β²*2

Term6: αβγ³δ = γ²*(αβγδ) = γ²*2

Term7: αβγδ³ = δ²*(αβγδ) = δ²*2

Similarly, the other terms:

Term2: α²β²γ² = (αβγ)^2 / α^something? Wait, but maybe express in terms of known variables.

Alternatively, note that Term2 is (αβ)^2 * γ²? Not sure. Wait Term2 is α²β²γ², which is (αβγ)^2. Similarly Term3 is α²β²δ², which is (αβδ)^2. Term5 is α²γ²δ² = (αγδ)^2. Term8 is β²γ²δ² = (βγδ)^2.

But since we know that αβγδ = 2, then:

For example, Term2: (αβγ)^2 = (αβγ)^2 = ?

But also, we know that αβγ + αβδ + αγδ + βγδ = 0 (from Vieta's). Also, each of these terms is (αβγ), etc.

But αβγ = (αβγδ)/δ = 2/δ. Similarly, αβδ = 2/γ, αγδ = 2/β, βγδ = 2/α. Therefore:

Term2: (αβγ)^2 = (2/δ)^2 = 4 / δ²

Similarly:

Term3: (αβδ)^2 = (2/γ)^2 = 4 / γ²

Term5: (αγδ)^2 = (2/β)^2 = 4 / β²

Term8: (βγδ)^2 = (2/α)^2 = 4 / α²

Therefore, terms 2, 3, 5, 8 can be written as 4/δ², 4/γ², 4/β², 4/α².

So let me rewrite the entire expression:

S1*S2*S3 = [2α² + 2β² + 2γ² + 2δ²] + [α²β²γ² + α²β²δ² + α²γ²δ² + β²γ²δ²]

Wait, no. Wait:

Wait, the first four terms:

Term1 + Term4 + Term6 + Term7 = 2α² + 2β² + 2γ² + 2δ²

And the remaining terms:

Term2 + Term3 + Term5 + Term8 = 4/α² + 4/β² + 4/γ² + 4/δ²

Therefore,

S1*S2*S3 = 2(α² + β² + γ² + δ²) + 4(1/α² + 1/β² + 1/γ² + 1/δ²)

We already know that α² + β² + γ² + δ² = P_2 = 4, so first term is 2*4 = 8.

Now, need to compute 1/α² + 1/β² + 1/γ² + 1/δ².

Note that 1/α² + 1/β² + 1/γ² + 1/δ² = (1/α + 1/β + 1/γ + 1/δ)^2 - 2(1/(αβ) + 1/(αγ) + 1/(αδ) + 1/(βγ) + 1/(βδ) + 1/(γδ))

Let me compute this.

First, compute sum of 1/α:

Let me denote Σ1/α. Since Σ1/α = (βγδ + αγδ + αβδ + αβγ)/αβγδ.

From Vieta's:

Σ1/α = (αβγ + αβδ + αγδ + βγδ)/αβγδ = S3 / S4 = 0 / 2 = 0.

Therefore, Σ1/α = 0.

Therefore, Σ1/α² = (Σ1/α)^2 - 2Σ(1/(αβ)) = 0 - 2*(Σ1/(αβ)).

So need to compute Σ1/(αβ).

Σ1/(αβ) is the sum over all pairs of 1/(αβ). Which is equal to [Σ(αγδ + αβδ + αβγ)] / (αβγδ) ?

Wait, let me think. For example, 1/(αβ) + 1/(αγ) + 1/(αδ) + 1/(βγ) + 1/(βδ) + 1/(γδ).

Alternatively, note that Σ1/(αβ) = [Σ(γδ)] / (αβγδ). Because each term 1/(αβ) = γδ / (αβγδ). Therefore:

Σ1/(αβ) = [γδ + αδ + αγ + βδ + βγ + αβ] / (αβγδ). Wait, but the numerator is Σγδ = sum of products two at a time, which is 0.

Wait, but Σγδ = 0. Therefore Σ1/(αβ) = 0 / 2 = 0.

Wait, so Σ1/(αβ) = 0. Therefore Σ1/α² = 0 - 2*0 = 0? That can't be right because if Σ1/α² = 0 then 4*0 = 0. But that would make S1*S2*S3 = 8 + 4*0 = 8? But then R = 8? Wait, but that seems possible. Wait, let me check this again.

Wait, Σ1/α² = (Σ1/α)^2 - 2Σ1/(αβ). We had Σ1/α = 0 and Σ1/(αβ) = [Σγδ]/(αβγδ) = 0/2 = 0. Therefore Σ1/α² = 0 - 2*0 = 0. So that would make the second part of S1*S2*S3 equal to 4*0 = 0. Therefore, total S1*S2*S3 = 8 + 0 = 8. Therefore R = 8. Therefore, the cubic equation is x³ - 8x - 8 = 0.

Wait, but let me confirm this calculation again step by step, because that seems surprising.

First, Σ1/α² = (Σ1/α)^2 - 2Σ1/(αβ).

Σ1/α is 0, as we calculated.

Σ1/(αβ) = sum over all pairs of 1/(αβ). Let me compute that.

Note that each term 1/(αβ) = γδ / (αβγδ). Since αβγδ = 2, so each term 1/(αβ) = γδ / 2. Therefore Σ1/(αβ) = (Σγδ)/2. But Σγδ is sum of products two at a time, which is 0. Hence Σ1/(αβ) = 0 / 2 = 0. Therefore yes, Σ1/α² = 0 - 2*0 = 0.

Hence, S1*S2*S3 = 2*(α² + β² + γ² + δ²) + 4*(Σ1/α²) = 2*4 + 4*0 = 8 + 0 = 8. Therefore R = 8.

So the cubic equation for S1, S2, S3 is x³ - 8x - 8 = 0.

Wait, but let me check if this is correct. If we have S1 + S2 + S3 = 0, S1*S2 + S1*S3 + S2*S3 = -8, and S1*S2*S3 = 8. Then the cubic equation would be x³ - (S1 + S2 + S3)x² + (S1S2 + S1S3 + S2S3)x - S1S2S3 = x³ - 0x² + (-8)x - 8 = x³ - 8x - 8 = 0.

So, solving x³ - 8x - 8 = 0.

Now, need to find the roots of this cubic equation.

Let me try rational roots. Possible rational roots are ±1, ±2, ±4, ±8.

Testing x=2: 8 - 16 -8 = -16 ≠ 0

x=-2: -8 + 16 -8 = 0. Oh! x=-2 is a root.

Therefore, we can factor x³ - 8x - 8 as (x + 2)(x² - 2x - 4).

Let me check:

(x + 2)(x² - 2x - 4) = x³ - 2x² - 4x + 2x² - 4x - 8 = x³ - 8x - 8. Yes, correct.

Therefore, the roots are x = -2 and roots of x² - 2x - 4 = 0. Solving x² - 2x - 4 = 0:

x = [2 ± sqrt(4 + 16)] / 2 = [2 ± sqrt(20)] / 2 = [2 ± 2sqrt(5)] / 2 = 1 ± sqrt(5).

Therefore, the three roots are -2, 1 + sqrt(5), and 1 - sqrt(5). Therefore, the unordered set {S1, S2, S3} is { -2, 1 + sqrt(5), 1 - sqrt(5) }

But wait, hold on. Let me check if that's correct. Since we had S1 + S2 + S3 = 0, and the roots are -2, 1 + sqrt(5), 1 - sqrt(5). Let me check the sum:

-2 + (1 + sqrt(5)) + (1 - sqrt(5)) = -2 + 1 + 1 + sqrt(5) - sqrt(5) = 0. Correct. And the product S1*S2*S3 = (-2)(1 + sqrt(5))(1 - sqrt(5)) = (-2)(1 - 5) = (-2)(-4) = 8. Which matches our earlier calculation. Also, the sum S1*S2 + S1*S3 + S2*S3 was -8. Let me check if with the roots, we have:

For roots a, b, c: a + b + c = 0, ab + ac + bc = -8, abc = 8.

Let me check with a=-2, b=1 + sqrt(5), c=1 - sqrt(5):

ab + ac + bc = (-2)(1 + sqrt(5)) + (-2)(1 - sqrt(5)) + (1 + sqrt(5))(1 - sqrt(5))

= -2 - 2sqrt(5) -2 + 2sqrt(5) + (1 - 5)

= (-4) + (-4) = -8. Correct.

Therefore, yes, the roots are indeed -2, 1 + sqrt(5), 1 - sqrt(5).

Therefore, the required unordered set is { -2, 1 + sqrt(5), 1 - sqrt(5) }.

But let me check if there is any mistake here. Because the question says that α₁, α₂, α₃, α₄ are distinct roots. But does that affect anything? Since the cubic equation has three distinct roots (since discriminant of x² - 2x - 4 is positive), so yes. Therefore, the set { -2, 1 + sqrt(5), 1 - sqrt(5) } is the answer.

But wait, let me check if there is another way to get the answer. For example, maybe I made a mistake in calculation of S1*S2*S3. Let me cross-verify.

Earlier, we had S1*S2*S3 = 8. But also, if S1, S2, S3 are the roots of x³ - 8x - 8 = 0, then yes. But maybe I can think of another approach.

Alternatively, perhaps using the fact that each S_i is of the form p + q where pq = 2. Therefore, each S_i satisfies t² - S_i t + 2 = 0. So for each S_i, S_i² - 2 = S_i * q, where q = 2 / p. Wait, not sure.

Alternatively, since each S_i = p + q, with pq = 2, then p and q are roots of x² - S_i x + 2 = 0. Therefore, for each S_i, the discriminant is S_i² - 8. Therefore, if the roots are real or complex. But since we have three S_i values, which are -2, 1 + sqrt(5), 1 - sqrt(5), then discriminant for S_i=-2 is (-2)^2 - 8 = 4 -8 = -4 < 0. So the corresponding p and q are complex conjugates. For S_i = 1 ± sqrt(5), discriminant is (1 ± sqrt(5))² - 8 = 1 ± 2 sqrt(5) + 5 - 8 = (6 ± 2 sqrt(5)) - 8 = -2 ± 2 sqrt(5).

Wait, for S_i = 1 + sqrt(5):

Discriminant is (1 + sqrt(5))² - 8 = 1 + 2 sqrt(5) + 5 - 8 = (6 + 2 sqrt(5)) - 8 = -2 + 2 sqrt(5) ≈ -2 + 4.47 ≈ 2.47 > 0. Therefore, real roots. Similarly for S_i = 1 - sqrt(5), discriminant is (1 - sqrt(5))² - 8 = 1 - 2 sqrt(5) + 5 -8 = (6 - 2 sqrt(5)) - 8 = -2 - 2 sqrt(5) < 0. Therefore, complex roots.

But since our original equation x⁴ + 2x³ + 2 = 0 has four roots, some of which could be real or complex. However, the question says they are distinct roots. However, regardless of their nature, the set {S_i} is as we found. Therefore, the answer is { -2, 1 + sqrt(5), 1 - sqrt(5) }.

But let me check another way. Let me assume that one of the S_i is -2. For example, suppose that S1 = αβ + γδ = -2. Let me check if that is possible.

If αβ + γδ = -2, and since αβ * γδ = 2, then we have that αβ and γδ are roots of the equation t² - (-2)t + 2 = 0 ⇒ t² + 2t + 2 = 0. The roots are t = [-2 ± sqrt(4 - 8)] / 2 = -1 ± i. Therefore, αβ and γδ are complex numbers. Similarly, for the other S_i's.

But if two of the S_i are 1 ± sqrt(5), which are real numbers, then that would mean that the pair products αβ, γδ are real numbers. Wait, but αβ is a product of two roots. If the equation has four roots, some of which could be complex.

But maybe there are two real roots and two complex roots? Or all complex roots? Let me check the original equation.

Original equation is x⁴ + 2x³ + 2 = 0. Let me check for real roots. Let me consider f(x) = x⁴ + 2x³ + 2. For x real:

- When x approaches ±∞, x⁴ dominates, so f(x) approaches +infty.

At x=0: f(0) = 0 + 0 + 2 = 2 > 0.

At x = -1: f(-1) = 1 - 2 + 2 = 1 > 0.

At x = -2: f(-2) = 16 - 16 + 2 = 2 > 0.

So f(x) is always positive? Wait, but that would imply that there are no real roots, which would mean all roots are complex. Therefore, since the equation is degree 4 with real coefficients, complex roots come in conjugate pairs. Therefore, there are two pairs of complex conjugate roots.

Therefore, the products αβ and γδ would be complex numbers, but their sum S1 = αβ + γδ. However, if αβ and γδ are conjugate pairs, then S1 would be real. Similarly for S2 and S3.

Wait, but according to our previous result, one of the S_i is -2, which is real, and others are 1 ± sqrt(5), which are also real. Therefore, all three S_i are real, which is possible because they are sums of complex conjugates.

But let me check if the equation x³ - 8x - 8 = 0 has three real roots? Well, we have one real root -2 and two other roots 1 ± sqrt(5). Wait, sqrt(5) is approx 2.236, so 1 + sqrt(5) ≈ 3.236, and 1 - sqrt(5) ≈ -1.236. Therefore, all three roots are real. Therefore, the set is { -2, 1 + sqrt(5), 1 - sqrt(5) }, which are three real numbers.

But if the original equation has four complex roots, how can the expressions S_i be real? Because they are sums of products of pairs of complex conjugate roots. For example, if α and β are complex conjugates, and γ and δ are complex conjugates, then αβ and γδ are real numbers, so their sum S1 is real. Similarly for S2 and S3. However, how do S2 and S3 end up being real?

Let me think. Suppose the roots are α, α*, γ, γ*, where * denotes complex conjugate.

Then S1 = αα* + γγ* = |α|² + |γ|², which is real.

S2 = αγ + α*γ*.

Similarly, S3 = αγ* + α*γ.

But S2 and S3 are complex numbers unless αγ is real. Wait, unless αγ is real or something else. Wait, if αγ is real, then S2 and S3 would be real? Let me think.

Suppose α and γ are complex numbers such that α = a + ib, γ = c + id. Then αγ = (a + ib)(c + id) = (ac - bd) + i(ad + bc). For αγ to be real, ad + bc = 0. Similarly, if α and γ are such that their product is real.

Alternatively, maybe not. However, regardless, S2 and S3 may not necessarily be real. However, in our case, we have derived that S2 and S3 are real. Therefore, this suggests that the expressions are real.

But maybe it's not necessary to check further. Given that we followed the algebra correctly and verified that the cubic equation has roots as derived, and that the symmetric sums check out, then the answer is { -2, 1 + sqrt(5), 1 - sqrt(5) }.

But let me check with an alternative approach. Suppose that the equation can be factored.

Original equation: x⁴ + 2x³ + 2 = 0.

Maybe try factoring into quadratics.

Assume that x⁴ + 2x³ + 2 = (x² + a x + b)(x² + c x + d)

Multiply out: x^4 + (a + c)x³ + (ac + b + d)x² + (ad + bc)x + b d

Set equal to x⁴ + 2x³ + 0x² + 0x + 2. Therefore:

1. a + c = 2

2. ac + b + d = 0

3. ad + bc = 0

4. b d = 2

So need to solve for a, b, c, d.

From 1: c = 2 - a

From 4: b d = 2. Let me assume that b and d are real numbers. Since the equation has no real roots (as we saw earlier), then b and d are positive or negative?

If b d = 2, they have the same sign. Let me suppose they are both positive. Let me attempt to solve.

From 3: a d + b c = 0. Substitute c = 2 - a:

a d + b (2 - a) = 0 ⇒ a d - 2b + a b = 0 ⇒ a (d + b) = 2b ⇒ a = (2b)/(d + b)

From equation 2: a c + b + d = 0. Substitute c = 2 - a:

a (2 - a) + b + d = 0.

Let me express this in terms of b and d. Since a = 2b/(d + b), let me substitute:

[2b/(d + b)]*(2 - 2b/(d + b)) + b + d = 0.

This looks complicated, but maybe assume some values for b and d. Since b d = 2. Let me suppose that b = d. Then b² = 2 ⇒ b = √2. So, b = d = √2. Then:

From equation 3: a d + b c = 0 ⇒ a * √2 + √2 * c = 0 ⇒ a + c = 0. But from equation 1: a + c = 2 ⇒ 0 = 2? Contradiction. Not possible.

Alternatively, assume that b = 1, d = 2. Then check equations:

From equation 3: a*2 + 1*c = 0 ⇒ 2a + c = 0. But from equation 1, c = 2 - a. Therefore:

2a + 2 - a = 0 ⇒ a + 2 = 0 ⇒ a = -2. Then c = 2 - (-2) = 4.

Check equation 2: a c + b + d = (-2)(4) + 1 + 2 = -8 + 3 = -5 ≠ 0. Not good.

Alternatively, try b = 2, d = 1:

From equation 3: a*1 + 2*c = 0 ⇒ a + 2c = 0. From equation 1: c = 2 - a. Substitute:

a + 2*(2 - a) = 0 ⇒ a + 4 - 2a = 0 ⇒ -a + 4 = 0 ⇒ a = 4. Then c = 2 - 4 = -2.

Check equation 2: a c + b + d = 4*(-2) + 2 + 1 = -8 + 3 = -5 ≠ 0. Again no good.

Alternatively, try b= -1, d=-2. Then b d = (-1)(-2) = 2.

Equation 3: a*(-2) + (-1)*c = 0 ⇒ -2a - c = 0 ⇒ c = -2a.

From equation 1: a + c = 2 ⇒ a - 2a = 2 ⇒ -a = 2 ⇒ a = -2 ⇒ c = -2*(-2) = 4.

Check equation 2: a c + b + d = (-2)(4) + (-1) + (-2) = -8 - 3 = -11 ≠ 0. Not good.

Alternatively, maybe b and d are not rational? Let me consider solving equations:

From above, b d = 2, and let me express a and c in terms of b and d.

From equation 3: a d + b c = 0, and c = 2 - a. Therefore, a d + b(2 - a) = 0 ⇒ a(d - b) + 2b = 0 ⇒ a = -2b / (d - b)

From equation 2: a(2 - a) + b + d = 0. Substitute a:

Let me denote a = -2b / (d - b). Let me denote that d = 2 / b (from b d = 2). Therefore:

a = -2b / ( (2 / b) - b ) = -2b / ( (2 - b²)/b ) = -2b * (b / (2 - b²)) ) = -2b² / (2 - b²)

Now substitute into equation 2:

a(2 - a) + b + d = 0

First compute 2 - a = 2 - [ -2b² / (2 - b²) ] = 2 + 2b² / (2 - b²) = [ 2(2 - b²) + 2b² ] / (2 - b²) ) = [4 - 2b² + 2b² ] / (2 - b²) ) = 4 / (2 - b²)

Now compute a(2 - a) = [ -2b² / (2 - b²) ] * [ 4 / (2 - b²) ] = -8b² / (2 - b²)^2

Then equation 2: -8b² / (2 - b²)^2 + b + d = 0. Since d = 2 / b:

=> -8b² / (2 - b²)^2 + b + 2 / b = 0

Multiply both sides by b(2 - b²)^2 to eliminate denominators:

-8b³ + b(2 - b²)^2 * b + 2(2 - b²)^2 = 0

Wait, this seems complicated. Let me compute term by term:

Let me denote term1 = -8b² / (2 - b²)^2 * b(2 - b²)^2 = -8b³

term2 = [b + 2/b] * b(2 - b²)^2 = b^2(2 - b²)^2 + 2(2 - b²)^2

Therefore, equation becomes:

-8b³ + b²(2 - b²)^2 + 2(2 - b²)^2 = 0

Factor out (2 - b²)^2:

[ b² + 2 ] (2 - b²)^2 - 8b³ = 0

Let me expand (2 - b²)^2 = 4 - 4b² + b⁴

So:

[ b² + 2 ] (4 - 4b² + b⁴) - 8b³ = 0

Let me compute this:

First, expand [ b² + 2 ] * [ b⁴ - 4b² + 4 ]:

= b²*(b⁴ - 4b² + 4) + 2*(b⁴ - 4b² + 4)

= b^6 - 4b^4 + 4b² + 2b^4 - 8b² + 8

= b^6 - 2b^4 - 4b² + 8

Then subtract 8b³:

Total expression: b^6 - 2b^4 - 4b² + 8 - 8b³ = b^6 - 2b^4 - 8b³ - 4b² + 8 = 0

This is a sixth-degree equation in b, which is extremely complex. This seems not helpful. Therefore, factoring might not be the way to go here.

Therefore, returning to our previous result, since we have derived that the set is { -2, 1 + sqrt(5), 1 - sqrt(5) }, and all the symmetric sums check out, and the cubic equation is correct, I think this is the correct answer.

Thus, the unordered set is { -2, 1 + √5, 1 - √5 }.

But let me confirm once more with another approach. Suppose we take specific values. Let me try to compute one of the S_i.

For example, suppose we can find the value of S1 = αβ + γδ.

From Vieta, we know that αβγδ = 2. Also, we have that S1 = αβ + γδ. Let me denote p = αβ, q = γδ. Then pq = 2, and S1 = p + q. So p and q are roots of t² - S1 t + 2 = 0.

Also, from Vieta's formulas, we know that α + β + γ + δ = -2. Let me denote that α + β = m, γ + δ = n. Then m + n = -2. Also, from Vieta's, we have αβ + γδ = S1, and also αβ + αγ + αδ + βγ + βδ + γδ = 0.

So, αβ + γδ + αγ + αδ + βγ + βδ = 0 ⇒ S1 + [αγ + αδ + βγ + βδ] = 0.

But αγ + αδ + βγ + βδ = α(γ + δ) + β(γ + δ) = (α + β)(γ + δ) = m n. So:

S1 + m n = 0.

But since m + n = -2, and m n = ?

Also, note that m = α + β, and n = γ + δ.

Also, the products:

αβ = p, γδ = q.

Additionally, the sum of α + β + γ + δ = m + n = -2.

Also, for the quadratic equations:

For α and β: x² - m x + p = 0

For γ and δ: x² - n x + q = 0

Since the original equation is x⁴ + 2x³ + 2 = 0. Therefore, the quartic can be written as (x² - m x + p)(x² - n x + q) = 0. Expanding this gives:

x⁴ - (m + n)x³ + (mn + p + q)x² - (mq + np)x + pq = 0.

Compare to original equation x⁴ + 2x³ + 0x² + 0x + 2 = 0. Therefore:

- (m + n) = 2 ⇒ m + n = -2 (which matches)

mn + p + q = 0 ⇒ mn + S1 = 0 (since p + q = S1)

- (mq + np) = 0 ⇒ mq + np = 0

pq = 2

From before, we have mn + S1 = 0. Since m + n = -2, we can express mn in terms of m or n. Let me denote mn + S1 = 0 ⇒ mn = -S1. Also, since m + n = -2, we can write mn = -S1.

Additionally, mq + np = 0. But since p = αβ, q = γδ. Also, q = 2/p from pq = 2. So substitute:

mq + np = m*(2/p) + n*p = 0 ⇒ (2m)/p + n p = 0.

But we can express n = -2 - m, so:

(2m)/p + (-2 - m)p = 0 ⇒ (2m)/p - 2p - m p = 0 ⇒ Multiply both sides by p:

2m - 2p² - m p² = 0 ⇒ 2m - 2p² - m p² = 0.

Also, from mn = -S1 and m + n = -2, so mn = -S1 ⇒ m(-2 - m) = -S1 ⇒ -2m - m² = -S1 ⇒ S1 = m² + 2m.

Therefore, S1 = m² + 2m.

But also, from the equation above: 2m - 2p² - m p² = 0.

But since p = αβ, and we know that αβ is a root of t² - m t + p = 0. Wait, but p is also related to m. Wait, this seems too convoluted.

Alternatively, since S1 = m² + 2m and also S1 = p + q = p + 2/p. Let me write:

S1 = p + 2/p = m² + 2m.

So we have p + 2/p = m² + 2m. Also, since p is a root of the quadratic x² - m x + p = 0, so p² - m p + p = 0? Wait, no. Wait, the quadratic equation is x² - m x + p = 0, so the roots are α and β. But p = αβ. So from quadratic equation, the product of roots is p, which is given. Also, sum of roots is m.

But I don't see a direct relation between p and m. However, we have that p is also related through the equation above.

But this seems too tangled. Maybe pick an S1 value and check consistency. For example, suppose S1 = -2. Then S1 = m² + 2m = -2 ⇒ m² + 2m + 2 = 0 ⇒ m = [-2 ± sqrt(4 - 8)]/2 = -1 ± i. Therefore, m is complex. Which makes sense, since if the roots are complex, then m = α + β is a complex number.

Then, p + 2/p = -2. Let me solve for p:

p + 2/p = -2 ⇒ p² + 2 = -2p ⇒ p² + 2p + 2 = 0 ⇒ p = [-2 ± sqrt(4 - 8)] / 2 = -1 ± i. Which matches with the roots of the quadratic x² - m x + p = 0. Indeed, since m is -1 + i or -1 - i, then the quadratic would be x² - m x + p = 0, with m and p both complex.

This seems consistent.

Similarly, if we take S1 = 1 + sqrt(5), then S1 = m² + 2m. Let me solve:

m² + 2m - S1 = 0 ⇒ m = [-2 ± sqrt(4 + 4S1)] / 2 = [-2 ± 2sqrt(1 + S1)] / 2 = -1 ± sqrt(1 + S1).

For S1 = 1 + sqrt(5), this gives m = -1 ± sqrt(1 + 1 + sqrt(5)) = -1 ± sqrt(2 + sqrt(5)). Similarly, for S1 = 1 - sqrt(5), m = -1 ± sqrt(1 + 1 - sqrt(5)) = -1 ± sqrt(2 - sqrt(5)).

But since our original equation has no real roots, these m would be complex or what? Wait, sqrt(2 + sqrt(5)) is real, because sqrt(5) ≈ 2.236, so 2 + sqrt(5) ≈ 4.236, so sqrt of that is ≈ 2.06, real. Similarly, sqrt(2 - sqrt(5)) is sqrt of 2 - 2.236 ≈ -0.236, which is imaginary. Therefore, for S1 = 1 - sqrt(5), m would involve sqrt of negative number, meaning m is complex. Which again is consistent.

Therefore, all these S_i values are consistent. Therefore, the answer is indeed { -2, 1 + sqrt(5), 1 - sqrt(5) }.

Thus, the unordered set is { -2, 1 + √5, 1 - √5 }.

**Final Answer**
\boxed{\{-2, 1 - \sqrt{5}, 1 + \sqrt{5}\}}

To determine the unordered set
$$
\{\alpha_1\alpha_2 + \alpha_3\alpha_4, \alpha_1\alpha_3 + \alpha_2\alpha_4, \alpha_1\alpha_4 + \alpha_2\alpha_3\}
$$
where $ \alpha_1, \alpha_2, \alpha_3, \alpha_4 $ are the four **distinct** roots of the polynomial equation
$$
x^4 + 2x^3 + 0x^2 + 0x + 2 = 0,
$$
we employ **Vieta's formulas** to simplify the situation.

---

### 1. **Apply Vieta's Formulas**

Let the roots be $ \alpha_1, \alpha_2, \alpha_3, \alpha_4 $. The quartic equation can be written as
$$
x^4 + 2x^3 + 2 = 0.
$$
This corresponds to standard coefficients:
- $ a = 2 $
- $ b = 0 $
- $ c = 0 $
- $ d = 2 $

From Vieta’s formulas for $ x^4 + a x^3 + b x^2 + c x + d = 0 $, we have:

- $ \alpha_1 + \alpha_2 + \alpha_3 + \alpha_4 = -a = -2 $
- $ \alpha_1\alpha_2 + \alpha_1\alpha_3 + \alpha_1\alpha_4 + \alpha_2\alpha_3 + \alpha_2\alpha_4 + \alpha_3\alpha_4 = b = 0 $
- $ \alpha_1\alpha_2\alpha_3 + \alpha_1\alpha_2\alpha_4 + \alpha_1\alpha_3\alpha_4 + \alpha_2\alpha_3\alpha_4 = -c = 0 $
- $ \alpha_1\alpha_2\alpha_3\alpha_4 = d = 2 $

---

### 2. **Target Expression**

We define:
$$
S_1 = \alpha_1\alpha_2 + \alpha_3\alpha_4, \quad
S_2 = \alpha_1\alpha_3 + \alpha_2\alpha_4, \quad
S_3 = \alpha_1\alpha_4 + \alpha_2\alpha_3
$$

Our goal is to compute the set $ \{S_1, S_2, S_3\} $.

---

### 3. **Symmetric Sum Relationships**

It's known that the sum of all two-product pairs of roots is $ \sum_{1 \leq i < j \leq 4} \alpha_i\alpha_j = 0 $.
Thus,
$$
S_1 + S_2 + S_3 = 0.
$$
So the sum of these three values is zero.

Let us denote:

- $ S = S_1 + S_2 + S_3 = 0 $
- $ Q = S_1S_2 + S_1S_3 + S_2S_3 $
- $ R = S_1S_2S_3 $

Then, the three values $ S_1, S_2, S_3 $ are roots of the cubic equation:
$$
x^3 - Sx^2 + Qx - R = x^3 - 8x - 8 = 0
$$

This cubic equation is found by using Newton’s identities after computing the power sums of the original quartic roots.

---

### 4. **Solve the Cubic Equation**

We solve:
$$
x^3 - 8x - 8 = 0
$$

Try $ x = -2 $:
$$
(-2)^3 - 8(-2) - 8 = -8 + 16 - 8 = 0
$$
So $ x = -2 $ is a root.

Factor:
$$
x^3 - 8x - 8 = (x + 2)(x^2 - 2x - 4)
$$

Solve the quadratic:
$$
x^2 - 2x - 4 = 0 \Rightarrow x = \frac{2 \pm \sqrt{4 + 16}}{2} = \frac{2 \pm \sqrt{20}}{2} = 1 \pm \sqrt{5}
$$

So the three roots of the cubic are:
$$
x = -2, \quad 1 + \sqrt{5}, \quad 1 - \sqrt{5}
$$

---

### 5. **Conclusion**

The unordered set $ \{S_1, S_2, S_3\} $ consists of these three values:

$$
\boxed{ \{-2, 1 - \sqrt{5}, 1 + \sqrt{5} \} }
$$

Solution:
Employing the elementary symmetric polynomials ($s_1 = \alpha_1+\alpha_2+\alpha_3+\alpha_4 = -2$, $s_2 = \alpha_1\alpha_2 + \alpha_1\alpha_3 + \alpha_1\alpha_4 + \alpha_2\alpha_3 + \alpha_2\alpha_4 + \alpha_3\alpha_4 = 0$, $s_3 = \alpha_1\alpha_2\alpha_3 + \alpha_2\alpha_3\alpha_4 + \alpha_3\alpha_4\alpha_1 + \alpha_4\alpha_1\alpha_2 = 0$, and $s_4 = \alpha_1\alpha_2\alpha_3\alpha_4 = 2$) we consider the polynomial \[
P(x) = (x-(\alpha_1\alpha_2+\alpha_3\alpha_4))(x-(\alpha_1\alpha_3+\alpha_2\alpha_4))(x-(\alpha_1\alpha_4+\alpha_2\alpha_3))
\]Because $P$ is symmetric with respect to $\alpha_1, \alpha_2, \alpha_3, \alpha_4$, we can express the coefficients of its expanded form in terms of the elementary symmetric polynomials.  We compute \begin{eqnarray*}
P(x) & = & x^3 - s_2x^2 + (s_3s_1-4s_4)x + (-s_3^2-s_4s_1^2+s_4s_2) \\
& = & x^3 - 8x - 8 \\
& = & (x+2)(x^2-2x-4)
\end{eqnarray*}The roots of $P(x)$ are $-2$ and $1 \pm \sqrt{5}$, so the answer is $\boxed{\{1\pm\sqrt{5},-2\}}.$

$\textbf{Remarks:}$ It is easy to find the coefficients of $x^2$ and $x$ by expansion, and the constant term can be computed without the complete expansion and decomposition of $(\alpha_1\alpha_2+\alpha_3\alpha_4)(\alpha_1\alpha_3+\alpha_2\alpha_4)(\alpha_1\alpha_4+\alpha_2\alpha_3)$ by noting that the only nonzero 6th degree expressions in $s_1, s_2, s_3,$ and $s_4$ are $s_1^6$ and $s_4s_1^2$. The general polynomial $P$ constructed here is called the cubic resolvent and arises in Galois theory.