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Rezolvati in R ecuatia: ([tex]x^{2}[/tex] + x) ([tex]x^{2}[/tex] + x - 3) +2 = 0


Răspuns :

Răspuns:

x₁=1;    x₂= -2

x₁ = [tex]\frac{\sqrt{5}-1 }{2}[/tex]    x₂ = - [tex]\frac{\sqrt{5}+1 }{2}[/tex]

Explicație pas cu pas:

Notez x²+x=y și avem ecuația:

y(y-3)+2 = 0 ⇔ y²-3y+2 = 0

Δ=9-8 = 1

y₁ = (3+1):2 = 2

y₂ = (3-1):2 = 1

Cazul 1: y=2 echivalent cu  x²+x= 2 ⇔  x²+x -2 = 0

Δ=1+8 = 9

x₁=(-1+3):2 = 1

x₂=(-1-3):2 = -2

Cazul 2: y=1 echivalent cu x²+x= 1 ⇔  x²+x -1 = 0

Δ=1+4 = 5

x₁=[tex]\frac{-1+\sqrt{5} }{2} = \frac{\sqrt{5}-1 }{2}[/tex]

x₂= [tex]\frac{-1-\sqrt{5} }{2} = - \frac{\sqrt{5}+1 }{2}[/tex]

[tex](x^2+x)(x^2+x-3) + 2=0[/tex]   desfacem parantezele:

[tex]x^4+x^3-3x^2+x^3+x^2-3x+2 = 0[/tex]

[tex]x^4+2x^3-2x^2-3x+2=0[/tex]  scriem [tex]2x^3=3x^3-x^3[/tex]

[tex]x^4-x^3+3x^3-2x^2-3x+2=0[/tex]

[tex]x^3(x-1)+3x^3-3x-(2x^2-2)=0[/tex]

[tex]x^3(x-1)+3x(x^2-1)-2(x^2-1)=0[/tex]

[tex]x^3(x-1)+(3x-2)(x^2-1)=0[/tex]

[tex]x^3(x-1)+(3x-2)(x-1)(x+1)=0[/tex]

[tex](x-1)[x^3+(3x-2)(x+1)]=0[/tex]

[tex](x-1)(x^3+3x^2+3x-2x-2)=0[/tex]

[tex](x-1)(x^3+3x^2+x-2)=0 \implies[/tex]

[tex]x-1=0 \implies x_1=1[/tex]

sau

[tex]x^3+3x^2+x-2=0[/tex]

Rezolvam a doua ecuatie:

[tex]x^3+2x^2+x^2+x-2=0[/tex]

[tex]x^2(x+2)+x^2+x-2=0[/tex]

[tex]x^2(x+2)+x^2+2x-x-2=0[/tex]

[tex]x^2(x+2)+x(x+2)-(x+2)=0[/tex]

[tex](x+2)(x^2+x-1)=0 \implies[/tex]

[tex]x+2=0 \implies x_2=-2[/tex]

sau

[tex]x^2+x-1=0[/tex]

Rezolvam a doua ecuatie:

[tex]\Delta = 1^2-4\cdot(-1)\cdot1=1+4=5[/tex]

[tex]x_3=\dfrac{-1+\sqrt\Delta}{2}=\dfrac{-1+\sqrt5}{2}[/tex]

[tex]x_4=\dfrac{-1-\sqrt\Delta}{2}=\dfrac{-1-\sqrt5}{2}[/tex]

Asadar, am obtinut 4 solutii:

x1 = 1

x2 = -2

x3 = (√5-1)/2

x4 = (-√5-1)/2