Question

Dada a equação x² + 12x + 4 = 0, determine a soma dos inversos das raízes.

Answer 1

Resposta:

Olá bom dia!

x² + 12x + 4 = 0

Coeficientes:

a = 1 ; b = 12 ; c = 4

Δ = b² - 4ac

Δ = 12² - 4(1)(4)

Δ = 144 - 16

Δ = 128

√Δ = √128 = $\sqrt{2*64} = \sqrt{2} *\sqrt{64} = 8\sqrt{2}$

x = (-b ± √Δ) / 2a

x = (-12 ± 8√2)/2

x = -6 ± 4√2

x' = -6 + 4√2

x" = -6 - 4√2

Os inversos são:

1/x' = $\frac{1}{-6+4\sqrt{2} }$

1/x" = $\frac{1}{-6-4\sqrt{2} }$

Logo:

(1/x') + (1/x") =

= $\frac{1}{-6+4\sqrt{2} } + \frac{1}{-6-4\sqrt{2} }$

= $\frac{-6-4\sqrt{2}+6-4\sqrt{2} }{36 + 24\sqrt{2}-24\sqrt{2}-16\sqrt{4} }$

= $\frac{-8\sqrt{2} }{36-16\sqrt{4} }$

=$\frac{-8\sqrt{2} }{36-16*2}$

= $\frac{-8\sqrt{2} }{36-32}$

= $\frac{-8\sqrt{2} }{4}$

=$-2\sqrt{2}$  

Resposta:

(1/x') + (1/x") = -2√2