Question

Rezolvati pleaseee
2-x-x^3=0

Answer 1

   
[tex]\displaystyle\\ 2-x-x^3=0 \text{Observam ca } x = 1 \text{ este solutie a ecuatiei.}\\ \text{Verificare: } 2 - 1 - 1^3 = 2 - 1 - 1 = 0\\ \text{Rescriem ecuatia:}\\ -x^3 - x + 2 = 0\\ \text{Organizam convenabil termenii astfel:} \\ 0x^2 = (x^2 - x^2)\\ -x = (x - 2x)\\ -x^3 + (x^2 - x^2) + (x - 2x) + 2 = 0\\ \text{Regrupam termenii:}\\ (-x^3 + x^2) + (-x^2 + x) + (- 2x + 2) = 0\\ \text{Dam factor comun.}\\ -x^2(x - 1) - x(x - 1) - 2(x - 1) = 0\\ (x - 1)(-x^2 -x - 2)= 0 [/tex]


[tex]\displaystyle\\ \text{Ecuatia are o radacina reala si 2 radacini complex-conjugate}\\\\ x - 1 = 0\\ \boxed{x_1 = 1}\\\\ -x^2 -x - 2)= 0~~~\Big|~\cdot (-1)\\ x^2 +x + 2)= 0\\\\ x_{23} = \frac{-b\pm \sqrt{b^2-4ac} }{2a}= \frac{-1\pm \sqrt{1-8}}{2}= \frac{-1\pm \sqrt{-7}}{2}=\frac{-1\pm i\sqrt{7}}{2} \\\\ \boxed{x_2 = \frac{-1 + i\sqrt{7}}{2}}\\\\ \boxed{x_3 = \frac{-1 - i\sqrt{7}}{2}} [/tex]

Answer 2

[tex]\it 2-x-x^3 =0 \Leftrightarrow -x^3-x+2=0|_{\cdot(-1)} \Leftrightarrow x^3+x-2=0 \Leftrightarrow \\\;\\ \\\;\\ \Leftrightarrow x^3+x-1-1=0 \Leftrightarrow (x^3-1)+(x-1) =0 \Leftrightarrow \\\;\\ \\\;\\ \Leftrightarrow (x-1)(x^2+x+1)+(x-1) =0 \Leftrightarrow (x-1)(x^2+x+2)=0\Leftrightarrow[/tex]


[tex]\it \Leftrightarrow \begin{cases} \it x-1=0 \\\;\\ \\\;\\ \it x^2+x+2=0\end{cases} [/tex]


[tex]\it x^2+x+2 = x^2+x+\dfrac{1}{4} +\dfrac{7}{4} =\left(x+\dfrac{1}{2}\right)^2+\dfrac{7}{4} \ \textgreater \ 0 \\\;\\ \\\;\\ Deci,\ ecua\c{t}ia\ \ x^2+x+2 = 0 \ \ nu\ admite\ solu\c{t}ii\ reale[/tex]

În concluzie, ecuația din enunț admite soluția reală unică x = 1.