Question

resolva no complexo a equação x²+2x+5=0

Answer 1

vamos lá

x² +2x +5=0

a=1
b=2
c=5

Δ=b²-4ac
Δ=2²-4(1)(5)
Δ=4-20
Δ=-16
√Δ= √16(-1)
√Δ=√16×√-1 ⇒√-1=√i²
√Δ = √16.√i²
√Δ=±4i

$x=- \frac{b\pm \sqrt{\triangle} }{2a} \\ \\ x= \frac{-2\pm 4i }{2} =2( \frac{-2\pm2i}{2} )=-2\pm2i \\ \\ x'=-2+2i \\ \\ x"=-2-2i$

Answer 2

Resolução da questão, veja:

Determinar Delta:

$\mathsf{\Delta=b^2-4~\cdot~a~\cdot~c}\\\\\\ \mathsf{\Delta= 2^2-4~\cdot~1~\cdot~5}\\\\\\ \mathsf{\Delta= 4-20}\\\\\\ \large\boxed{\boxed{\mathsf{\Delta=-16.}}}$

Fórmula de Bhaskara:

$\mathsf{x = \dfrac{-2~\pm~\sqrt{-16}}{2~\cdot~1}}\\\\\\ \mathsf{x = \dfrac{-2~\pm~\sqrt{-16~\cdot~(-1)}}{2}}\\\\\\\ \mathsf{x = \dfrac{-2~\pm~\sqrt{16}}{2}}\\\\\\ \mathsf{x = \dfrac{-2~\pm~4i}{2}}\\\\\\ \large\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\mathsf{x = -1~\pm~2i}}}}}}}}}}}}}~\\\\\\\\ \textsf{Dando~o~conjunto~solu\c{c}\~ao}: ~\large\boxed{\boxed{\boxed{\boxed{\mathsf{S = \big\{-1+2i~,-1-2i\big\}}}}}}}}}}}$

Espero que te ajude. :-)

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