• Matéria: Matemática
  • Autor: Julianoc5680
  • Perguntado 5 anos atrás

Resolva as equaçôes
De 2° grau com uma incógnita
a) x² - 1 = 0

b) 3x² - 27 = 0

c) 5x² - 45x = 0

Respostas

respondido por: Anônimo
2

Resposta:

OLÁ

VAMOS A SUA PERGUNTA:⇒⇒

A:)→

x^{2} -1=0

x=\dfrac{0\pm\sqrt{0^2-4(-1)} }{2}

x=\dfrac{0\pm\sqrt{-4(-1)} }{2}

x=\dfrac{0\pm\sqrt{4} }{2}

x=\dfrac{0\pm2 }{2}

x=1

x=-1

\boxed{\bold{\displaystyle{\clubsuit\ \spadesuit\ \maltese\ x=1}}}\ \checkmark\\\\\boxed{\bold{\displaystyle{\clubsuit\ \spadesuit\ \maltese\ x=-1}}}\ \checkmark←↑ RESPOSTA.

B:)→

3x^2-27=0

x=\dfrac{0\pm\sqrt{0^2-4\times3(-27)} }{2\times3}

x=\dfrac{0\pm\sqrt{-4\times3(-27)} }{2\times3}

x=\dfrac{0\pm\sqrt{-12(-27)} }{2\times3}

x=\dfrac{0\pm\sqrt{324} }{2\times3}

x=\dfrac{0\pm18 }{2\times3}

x=\dfrac{0\pm18 }{6}

x=3

x=-3

\boxed{\bold{\displaystyle{\clubsuit\ \spadesuit\ \maltese\ x=3 }}}\ \checkmark\\\\\boxed{\bold{\displaystyle{\clubsuit\ \spadesuit\ \maltese\ x=-3}}}\ \checkmark← ↑RESPOSTA.

C:)→

5x^2-45x=0

x=\dfrac{-(-45)\pm\sqrt{(-45)^2} }{2\times5}

x=\dfrac{-(-45)\pm45 }{2\times5}

x=\dfrac{45\pm45 }{10}

x=\dfrac{90}{10}

x=9

x=\dfrac{0}{10}

x=0

\boxed{\bold{\displaystyle{\clubsuit\ \spadesuit\ \maltese\ x=9 }}}\ \checkmark\\\\\boxed{\bold{\displaystyle{\clubsuit\ \spadesuit\ \maltese\ x=0}}}\ \checkmark←↑ RESPOSTA.

Explicação passo-a-passo:

ESPERO TER AJUDADO

Anexos:
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