Question

Am si eu nevoie de ajutor pentru a rezolva ecuatia asta: radical de ordinul trei din x la a 3 a -2*x la a 2 a +2x+7=x+1

Answer 1

      $\sqrt[3]{x^{3}-2 x^{2} +2x+7} = x+1$ ⇔
⇔  $x^{3}-2 x^{2} +2x+7 = (x+1)^{3}$ ⇔
 
$(x+y)^{3}= x^{3} +3 x^{2} y+3x y^{2} + y^{3}$ (formula)

⇔  $x^{3}-2 x^{2} +2x+7=x^{3}+3 x^{2} +3x+1$ ⇔
⇔  $x^{3}-2 x^{2} +2x+7-x^{3}-3 x^{2} -3x-1=0$ ⇔
⇔  $-5 x^{2} -x+6=0$

Δ=$b^{2} -4ac= (-1)^{2} -4*(-5)*6=1+120=121$

[tex] x_{1}= \frac{1+ 11 }{2*(-5)}= \frac{12}{-10} =- \frac{6}{5} [/tex]
$x_{2} = \frac{1-11}{2*(-5)}= \frac{-10}{-10}=1$ ⇒

⇒ $x_{1}= -\frac{6}{5}$ si $x_{2} = 1$.