• Matéria: Matemática
  • Autor: carol897228
  • Perguntado 6 anos atrás


\frac{x {}^{2} }{5} - \frac{x}{3} = \frac{2}{15}
bhaskara equação de 2° grau​

Respostas

respondido por: dougOcara
0

Resposta:

Explicação passo-a-passo:

\displaystyle \frac{x^{2}}{5} - \frac{x}{3} = \frac{2}{15}

mmc(3,5,15)

3, 5, 15 | 3

1, 5, 5 | 5

1, 1, 1 | 1

mmc(3,5,15) = 3.5=15

\displaystyle \frac{3x^2-5x}{15} =\frac{2}{15}\\\\3x^2-5x=2\\3x^2-5x-2=0\\\\\displaystyle Aplicando~a~f\'{o}rmula~de~Bhaskara~para~3x^{2}-5x-2=0~~\\e~comparando~com~(a)x^{2}+(b)x+(c)=0,~temos~a=3{;}~b=-5~e~c=-2\\\\\Delta=(b)^{2}-4(a)(c)=(-5)^{2}-4(3)(-2)=25-(-24)=49\\\\x^{'}=\frac{-(b)-\sqrt{\Delta}}{2(a)}=\frac{-(-5)-\sqrt{49}}{2(3)}=\frac{5-7}{6}=\frac{-2}{6}=-\frac{1}{3}\\\\x^{''}=\frac{-(b)+\sqrt{\Delta}}{2(a)}=\frac{-(-5)+\sqrt{49}}{2(3)}=\frac{5+7}{6}=\frac{12}{6}=2\\\\S=\{-\frac{1}{3},~2\}

Perguntas similares