Răspuns:
Explicație pas cu pas:
a) metoda 1
Observam ca :
5X² - 3X - 2 = 5x² - 5X + 2X - 2= 5X*(X - 1) +2*(X - 1) = (X - 1)(5X + 2)
Asadar
(X - 1)(5X + 2) = 0
X - 1 = 0 ⇒ X = 1
sau
5X + 2 = 0 ⇒ X = -2/5
metoda 2
[tex]x_{1} = \frac{-(-3)+\sqrt{(-3)^2-4*5*(-2)} }{2*5} = \frac{3+\sqrt{9+40} }{10} = \frac{3+\sqrt{49} }{10} = \frac{3+7 }{10} = \frac{10}{10} = 1\\x_{2} = \frac{-(-3)-\sqrt{(-3)^2-4*5*(-2)} }{2*5} = \frac{3-\sqrt{9+40} }{10} = \frac{3-\sqrt{49} }{10} = \frac{3-7 }{10} = \frac{-4}{10} = -\frac{2}{5}[/tex]
b) X² - X = X + 2
X² - 2X - 2 = 0
metoda 1
observam ca X² - 2X - 2 = X² - 2X +1 -3 = (X - 1)² -3
Asadar
(X - 1)² -3 = 0
(X - 1)² = 3 = (√3)²
X - 1 = √3 ⇒X = 1 + √3
sau
X - 1 = -√3 ⇒ X = 1 - √3
metoda 2
[tex]x_{1} = \frac{-(-2)+\sqrt{(-2)^2-4*1*(-2)} }{2*1} = \frac{2+\sqrt{4+8} }{2} = \frac{2+\sqrt{12} }{2} = \frac{2+\sqrt{4*3}}{2} = \frac{2+\sqrt{4}*\sqrt{3} }{2} = \frac{2+2*\sqrt{3} }{2} = 1+\sqrt{3}[/tex]
[tex]x_{2} = \frac{-(-2)-\sqrt{(-2)^2-4*1*(-2)} }{2*1} = \frac{2-\sqrt{4+8} }{2} = \frac{2-\sqrt{12} }{2} = \frac{2-\sqrt{4*3}}{2} = \frac{2-\sqrt{4}*\sqrt{3} }{2} = \frac{2-2*\sqrt{3} }{2} = 1-\sqrt{3}[/tex]
