• Matéria: Matemática
  • Autor: anaclarapereir7281
  • Perguntado 3 anos atrás

(x+3) elevado a 2 + (x+3) elevado a dois-116=0

Respostas

respondido por: cavalcantidnl
0

Resposta:

\left(x+3\right)^2=58

\left(x+3\right)^2+\left(x+3\right)^2-116=0

\left(x+3\right)^2+\left(x+3\right)^2-116+116=0+116

\left(x+3\right)^2+\left(x+3\right)^2=116

2\left(x+3\right)^2=116

\frac{2\left(x+3\right)^2}{2}=\frac{116}{2}

\left(x+3\right)^2=58

x=\sqrt{58}-3,\:x=-\sqrt{58}-3

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