Question

Am nevoie de ajutor, va rog...
Rezolvati in R ecuatia: ($x^{2}$ + x) ($x^{2}$ + x - 3) +2 = 0

Answer 1

Răspuns:

x₁=1;    x₂= -2

x₁ = $\frac{\sqrt{5}-1 }{2}$    x₂ = - $\frac{\sqrt{5}+1 }{2}$

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=$\frac{-1+\sqrt{5} }{2} = \frac{\sqrt{5}-1 }{2}$

x₂= $\frac{-1-\sqrt{5} }{2} = - \frac{\sqrt{5}+1 }{2}$

Answer 2

$(x^2+x)(x^2+x-3) + 2=0$   desfacem parantezele:

$x^4+x^3-3x^2+x^3+x^2-3x+2 = 0$

$x^4+2x^3-2x^2-3x+2=0$  scriem $2x^3=3x^3-x^3$

$x^4-x^3+3x^3-2x^2-3x+2=0$

$x^3(x-1)+3x^3-3x-(2x^2-2)=0$

$x^3(x-1)+3x(x^2-1)-2(x^2-1)=0$

$x^3(x-1)+(3x-2)(x^2-1)=0$

$x^3(x-1)+(3x-2)(x-1)(x+1)=0$

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

$(x-1)(x^3+3x^2+3x-2x-2)=0$

$(x-1)(x^3+3x^2+x-2)=0 \implies$

$x-1=0 \implies x_1=1$

sau

$x^3+3x^2+x-2=0$

Rezolvam a doua ecuatie:

$x^3+2x^2+x^2+x-2=0$

$x^2(x+2)+x^2+x-2=0$

$x^2(x+2)+x^2+2x-x-2=0$

$x^2(x+2)+x(x+2)-(x+2)=0$

$(x+2)(x^2+x-1)=0 \implies$

$x+2=0 \implies x_2=-2$

sau

$x^2+x-1=0$

Rezolvam a doua ecuatie:

$\Delta = 1^2-4\cdot(-1)\cdot1=1+4=5$

$x_3=\dfrac{-1+\sqrt\Delta}{2}=\dfrac{-1+\sqrt5}{2}$

$x_4=\dfrac{-1-\sqrt\Delta}{2}=\dfrac{-1-\sqrt5}{2}$

Asadar, am obtinut 4 solutii:

x1 = 1

x2 = -2

x3 = (√5-1)/2

x4 = (-√5-1)/2