Question

como resolver x^+x(x-6)=0

Answer 1

Resposta:

$S=\{0,3\}$

Explicação passo-a-passo:

x² + x (x - 6) = 0

x² + x² - 6x = 0

2x² - 6x = 0

a = 2

b = (- 6)

c = 0

$\Delta=b^{2} -4ac\\\Delta=(-6)^{2} -4\times2\times0\\\Delta=36-0\\\Delta=36$

$x=\frac{-b\pm\sqrt{\Delta} }{2a} \\\\x=\frac{-(-6)\pm\sqrt{36} }{2\times2}\\ \\x=\frac{6\pm6}{4}$

$x_{1} =\frac{6+6}{4} =\frac{12}{4} =3\\\\x_{2} =\frac{6-6}{4} =\frac{0}{4} =0$

$S=\{0,3\}$