Matemática, perguntado por aaaaaa2360, 7 meses atrás

Observe os exemplos resolvidos acima e resolva as seguintes equações do 2º grau.
a) x2 - 49 = 0
b) 2x2 - 50 = 0
c) 72-7=0
d) 5x2 - 15 = 0
e) 21 = 7x2
f) 5x2 + 20 = 0​

Soluções para a tarefa

Respondido por crquadros
2

Resposta:

a)\ \boxed{x=\pm7}\\\\b)\ \boxed{x=\pm5}\\\\c)\ \boxed{x=\pm1}\\\\d)\ \boxed{x=\pm\sqrt{3}}\\\\e)\ \boxed{x=\pm\sqrt{3}}\\\\f)\ \boxed{x=\pm\sqrt{-4}} \Longleftarrow \ \ \in \ n\'{u}meros\ Irracionais\\

Explicação passo-a-passo:

Vamos resolver os exercícios isolando a variável "x":

a)\ x^2-49=0\\.\ \ \ x^2=49\\.\ \ \ x=\sqrt{49} = \pm7 \longrightarrow \boxed{x=\bf{\pm7}}\\\\b)\ 2x^2-50=0\\.\ \ \ 2x^2=50\\\\.\ \ \ x^2=\dfrac{50}{2}\\\\.\ \ \ x^2=25\\.\ \ \ x=\sqrt{25} = \pm5 \longrightarrow \boxed{x=\bf{\pm5}}

c)\ 7x^2-7=0\\.\ \ \ 7x^2=7\\\\.\ \ \ x^2=\dfrac{7}{7}\\\\.\ \ \ x^2=1\\.\ \ \ x=\sqrt{1} = \pm1 \longrightarrow \boxed{x=\bf{\pm1}}\\\\d)\ 5x^2-15=0\\.\ \ \ 5x^2=15\\\\.\ \ \ x^2=\dfrac{15}{5}\\\\.\ \ \ x^2=3\\.\ \ \ x=\sqrt{3} \longrightarrow \boxed{x=\bf{\pm\sqrt{3}}}

e)\ 21=7x^2\\\\.\ \ \ x^2=\dfrac{21}{7}\\\\.\ \ \ x^2=3\\\\.\ \ \ x=\sqrt{3} \longrightarrow \boxed{x=\bf{\pm\sqrt{3}}}\\\\f)\ 5x^2+20=0\\.\ \ \ 5x^2=-20\\\\.\ \ \ x^2=\dfrac{-20}{5}\\\\.\ \ \ x^2=-4\\.\ \ \ x=\sqrt{-4} \longrightarrow \boxed{x=\bf{\pm\sqrt{-4}}} \Longleftarrow Solu\c{c}\~{a}o\ pertence\ aos\ n\'umeros\ Irracionais

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grazymoreninha1: cadê o restante?
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