Question

(FEI) A equação (x+1)²/x² + 5(x+1/x) + 6=0 admite 2 raizes reais cujas soma é? (ajudem-me por favor!)

Answer 1

$Resolu\c{c}\~ao \to \left\{\begin{array}{ccc}\frac{(x+1)^2}{x^2}+5*(\frac{x+1}{x})+6=0\\\\ \frac{(x^2+2x+1)}{x^2}+(\frac{5x+5}{x})+6 = 0\\\\\frac{x*(x^2+2x+1)+x^2*(5x+5)+6x^3=0}{x^3}\\\\x^3+2x^2+x+5x^3+5x^2+6x^3=0\\\\12x^3+7x^2+x=0\\\\x*(12x^2+7x+1)=0\\\\\boxed{\boxed{x=0}}\\\\12x^2+7x+1=0\\\\\boxed{\boxed{S = -\frac{b}{a} = -\frac{7}{12}}}\end{array}\right$

Espero ter ajudado. =^.^=

Answer 2

 Outra!

 Consideremos $\frac{(x+1)}{x}=\lambda$; segue que,

$\lambda^2+5\lambda+6=0\\\\(\lambda+2)(\lambda+3)=0\\\\\boxed{\lambda=-2}\\\boxed{\lambda=-3}$
 
 Substituindo esses valores na condição considerada/sugerida, temos:

Quando $\lambda=-2$:

$\frac{x+1}{x}=-2\\\\x+1=-2x\\\\3x=-1\\\\\boxed{x=-\frac{1}{3}}$


Quando $\lambda=-3$:

$\frac{x+1}{x}=-3\\\\x+1=-3x\\\\4x=-1\\\\\boxed{x=-\frac{1}{4}}$
 
 Somemos,

$x'+x''=\\\\-\frac{1}{3/4}-\frac{1}{4/3}=\\\\\frac{-4-3}{12}=\\\\\boxed{\boxed{-\frac{7}{12}}}$