Question

10 -Se m e n são as raízes das equação 3x² +x -1 = 0, então m/n + n/m , é: *

a)7/3
b)-7/3
c)3/7
d)-3/7

Answer 1

$\displaystyle \sf 3x^2+x-1 = 0 \\\\\ ra{\'i}}zes : \ m \ e \ n \\\\\\ soma\ das\ ra{\'i}}zes : \\\\ m+n =\frac{-b}{a} \to \boxed{\sf m+n = \frac{-1}{3} }\\\\\\ Produto \ das\ ra{\'i}}zes : \\\\ m\cdot n = \frac{c}{a} \to \boxed{\sf m\cdot n =\frac{-1}{3} }$

Temos :

$\displaystyle \sf \frac{m}{n}+\frac{n}{m}=\ ? \\\\ Tirando \ o \ MMC : \\\\ \frac{m}{n/m}+\frac{n}{m/n} \\\\\\ \frac{m^2+n^2}{m\cdot n } \\\\\\ \frac{m^2+n^2+2\cdot m\cdot n - 2\cdot m\cdot n }{m\cdot n } \\\\\\ \frac{(m+n)^2-2\cdot m\cdot n}{m\cdot n}$

Substituindo os respectivos valores :

$\displaystyle \sf \frac{(m+n)^2-2\cdot m\cdot n}{m\cdot n} \\\\\\ \frac{\displaystyle \left(\frac{-1}{3}\right)^2-2\cdot \left(\frac{-1}{3}\right)}{\displaystyle \left (\frac{-1}{3}\right ) } \\\\\\ \displaystyle \frac{-1}{3} - 2 = \frac{-1-2\cdot 3}{3}= \frac{-1-6}{3}=\frac{-7}{3} \\\\\\ Portanto : \\\\ \huge\boxed{\sf \frac{m}{n}+\frac{n}{m}=\frac{-7}{3} }\checkmark$

letra b

Answer 2

Resposta:

Olá bom dia!

Calculando as raízes por Bhaskara:

3x² + x - 1 = 0

a = 3 ; b = 1 ; c = -1

Δ = b² - 4ac

Δ = 1² - 4(3)(-1)

Δ = 1 + 12

Δ = 13

√Δ = √13 = 3,60

x = (-b ±√Δ) / 2a

x = (-1 ± 3,60) / 6

x' = m = (-1 + 3,60) / 6

x' = m = 0,43

x" = n = (-1 - 3,60) / 6

x" = n = -0,76

m/n = 0,43/-0,76 = -0,56

n/m = -0,76/0,43 = - 1,76

m/n + n/m = -1,76 + (-0,56) = -2,32 =~-7/3

Alternativa B