• Matéria: Matemática
  • Autor: panelasavc7473
  • Perguntado 4 anos atrás

resolva as equações abaixo

a)x²+2x-4x=0

b) x²+6x=0

c)x²+2x=8x

d)x²=3x

e)x²-7x=0

f)x²+9x=0

g) x²-3x=-5x

me ajudaaaaaa​

Respostas

respondido por: ArthurCMaurer
1

Fórmula de Bhaskara: \frac{-b\pm\sqrt\Delta}{2a}

Fórmula de Delta: \Delta=b^2-4ac

Soma e produto (Relações de Girard): \left \{ {{r_1+r_2=\frac{-b}{a}} \atop {r_1.r_2=\frac{c}{a}}} \right.

Resposta:

a) x^2-2x=0\\\\\Delta=(-2)^2-4(1)(0)\\=4-0=4\\\\x=\frac{2\pm\sqrt4}{2}=\frac{2\pm2}{2}\\x'=\frac{2+2}{2}=2\\x''=\frac{2-2}{2}=0     b) x^2+6x=0\\\\\Delta=6^2-4(1)(0)\\=36-0=36\\\\x=\frac{-6\pm\sqrt{36}}{2}=\frac{-6\pm6}{2}\\x'=\frac{-6+6}{2}=0\\x''=\frac{-6-6}{2}=-6     c) x^2-6x=0\\\\\Delta=(-6)^2-4(1)(0)\\=36-0\\\\x=\frac{6\pm\sqrt{36}}{2}=\frac{6\pm6}{2}\\x'=\frac{6+6}{2}=6\\x''=\frac{6-6}{2}=0

d) x^2-3x=0\\\\\Delta=(-3)^2-4(1)(0)\\=9-0=9\\\\x=\frac{3\pm\sqrt9}{2}=\frac{3\pm3}{2}\\x'=\frac{3+3}{2}=3\\x''=\frac{3-3}{2}=0     e) x^2-7x=0\\\\\Delta=(-7)^2-4(1)(0)\\=49-0=49\\\\x=\frac{7\pm\sqrt{49}}{2}=\frac{7\pm7}{2}\\x'=\frac{7+7}{2}=7\\x''=\frac{7-7}{2}=0     f) x^2+9x=0\\\\\Delta=9^2-4(1).0\\=81-0=81\\\\x=\frac{-9\pm\sqrt{81}}{2}=\frac{-9\pm9}{2}\\x'=\frac{-9+9}{2}=0\\x''=\frac{-9-9}{2}=-9

g) x^2+2x=0\\\\\Delta=2^2-4(1)(0)\\=4-0=4\\\\x=\frac{-2\pm\sqrt4}{2}=\frac{-2\pm2}{2}\\x'=\frac{-2+2}{2}=0\\x''=\frac{-2-2}{2}=-2


ArthurCMaurer: Pronto, acho que está certo
panelasavc7473: obrigada mano ajudou muito
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