Question

2. Determina el conjunto solución para XER
b) 2x^4 - 3x^3 + 5x^2 + 3= 7​

Answer 1

Respuesta:

2x⁴ - 3x³ + 5x² + 3 = 7

2x⁴ - 3x³ + 5x² - 4 = 0

2x⁴ - (x³ + 2x³) + (4x² + x²) + (4x - 4x) - 4 = 0

2x⁴ - x³ - 2x³ + 4x² + x² + 4x - 4x - 4 = 0

2x⁴ - x³ + 4x² + 4x - 2x³ + x² - 4x - 4 = 0

(2x⁴ - x³ + 4x² + 4x) - (2x³ - x² + 4x + 4) = 0

x.(2x³ - x² + 4x + 4) - (2x³ - x² + 4x + 4) = 0

(x - 1).(2x³ - x² + 4x + 4) = 0

First case: (x - 1) = 0 → x - 1 = 0 → x = 1

Second case: (2x³ - x² + 4x + 4) = 0

2x³ - x² + 4x + 4 = 0 → it's necessary to eliminate the term at the power 2

2x³ - x² + 4x + 4 = 0 → let: x = z + (1/6)

2.[z + (1/6)]³ - [z + (1/6)]² + 4.[z + (1/6)] + 4 = 0

2.[z + (1/6)]².[z + (1/6)] - [z² + (1/3).z + (1/36)] + 4z + (2/3) + 4 = 0

2.[z² + (1/3).z + (1/36)].[z + (1/6)] - z² - (1/3).z - (1/36) + 4z + (14/3) = 0

2.[z³ + (1/6).z² + (1/3).z² + (1/18).z + (1/36).z + (1/216)] - z² + (11/3).z + (167/36) = 0

2.[z³ + (1/2).z² + (1/12).z + (1/216)] - z² + (11/3).z + (167/36) = 0

2z³ + z² + (1/6).z + (1/108) - z² + (11/3).z + (167/36) = 0

2z³ + (23/6).z + (251/54) = 0 ← no term at power 2

2z³ + (23/6).z + (251/54) = 0 → let: z = u + v

2.(u + v)³ + (23/6).(u + v) + (251/54) = 0

2.[(u + v)².(u + v)] + (23/6).(u + v) + (251/54) = 0

2.[(u² + 2uv + v²).(u + v)] + (23/6).(u + v) + (251/54) = 0

2.[u³ + u²v + 2u²v + 2uv² + uv² + v³] + (23/6).(u + v) + (251/54) = 0

2.[u³ + v³ + 3u²v + 3uv²] + (23/6).(u + v) + (251/54) = 0

2.[(u³ + v³) + (3u²v + 3uv²)] + (23/6).(u + v) + (251/54) = 0

2.[(u³ + v³) + 3uv.(u + v)] + (23/6).(u + v) + (251/54) = 0

2.(u³ + v³) + 6uv.(u + v) + (23/6).(u + v) + (251/54) = 0 → you can factorize (u + v)

2.(u³ + v³) + (u + v).[6uv + (23/6)] + (251/54) = 0 → suppose that: [6uv + (23/6)] = 0 ← equation (1)

2.(u³ + v³) + (u + v).[0] + (251/54) = 0

2.(u³ + v³) + (251/54) = 0 ← equation (2)

You can get a system of 2 equations:

(1) : [6uv + (23/6)] = 0

(1) : 6uv = - 23/6

(1) : uv = - 23/36

(1) : u³v³ = - (23/36)³

(2) : 2.(u³ + v³) + (251/54) = 0

(2) : 2.(u³ + v³) = - 251/54

(2) : u³ + v³ = - 251/108

Let: U = u³

Let: V = v³

You can get a new system of 2 equations:

(1) : UV = - (23/36)³ ← this is the product P

(2) : U + V = - 251/108 ← this is the sum S

You know that the values S & P are the solutions of the following equation:

x² - Sx + P = 0

x² + (251/108).x - (23/36)³ = 0

Δ = (251/108)² + [4 * (23/36)³]

Δ = (1/3)² * 58

x₁ = [- (251/108) - (1/3).√58]/2 = - (251/216) - (1/6).√58 ← this is U

x₂ = [- (251/108) + (1/3).√58]/2 = - (251/216) + (1/6).√58 ← this is V

Recall: u³ = U → u = U^(1/3) = [- (251/216) - (1/6).√58]^(1/3)

Recall: v³ = V → v = V^(1/3) = [- (251/216) + (1/6).√58]^(1/3)

Recall: z = u + v

Recall: x = z + (1/6)

x = u + v + (1/6)

x = [- (251/216) - (1/6).√58]^(1/3) + [- (251/216) + (1/6).√58]^(1/3) + (1/6)

x ≈ [- 1.34466712872526] + [0.475127914738685] + (1/6)

x ≈ - 0.702872547319909

→ Solution = { - 0.702872547319909 ; 1 }

Explicación paso a paso:

Espero que te sirva :3