• Matéria: Matemática
  • Autor: Isa1888
  • Perguntado 3 anos atrás

4x² + 20x +221=0
a) -5/2+7i
b) -5/2-7i
c) -5/2±7i
d) 5±7i
e) 5/2±7i/2

Respostas

respondido por: dougOcara
1

Resposta:

-5/2±√7i

Explicação passo a passo:

\displaystyle Aplicando~a~f\acute{o}rmula~de~Bhaskara~para~4x^{2}+20x+221=0~~e~comparando~com~(a)x^{2}+(b)x+(c)=0,~determinamos~os~coeficientes:~\\a=4{;}~b=20~e~c=221\\C\acute{a}lculo~do~discriminante~(\Delta):&\\&~\Delta=(b)^{2}-4(a)(c)=(20)^{2}-4(4)(221)=400-(3536)=-3136\\\sqrt{\Delta}=\sqrt{-3136}= \sqrt{-2^6.7}=\sqrt{2^6}.\sqrt{7}.\sqrt{-1}=2^3\sqrt{7}i=8\sqrt{7}i\displaystyle C\acute{a}lculo~das~raizes:&\\x^{'}=\frac{-(b)-\sqrt{\Delta}}{2(a)}=\frac{-(20)-8\sqrt{7}i}{2(4)}=-\frac{5}{2}-\sqrt{7}i\\\\x^{''}=\frac{-(b)+\sqrt{\Delta}}{2(a)}=\frac{-(20)+8\sqrt{7}i}{2(4)}=-\frac{5}{2}+\sqrt{7}i


Isa1888: obrigada
Isa1888: https://brainly.com.br/tarefa/46340870 se puder me ajudar nessa
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