Question

6) Aplicando a fórmula de Bhaskara, resolva as seguintes equações do 2º grau. a) 3x² – 7x + 4 = 0 b) 9y² – 12y + 4 = 0 c) 5x² + 3x + 5 = 0

Answer 1

a) 3x^2-7x+4=0

x= -b+ou-√b^2-4ac/2a

x= 7+ou-√49-4*3*4/2*3
x'= 7+√49-48/6
x'= 7+√1/6
x'= 7+1/6
x'=8/6= 4/3
x"= 7-1/6= 6/6
x"=1
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b) 9y2-12y+4=0
y= 12+ou-√144-144/18
y=12+ ou -√0/18
y'= 12+0/18
y'= 12/18
y'= 2/3
y"=12-0/18
y"=12/18
y"=2/3
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c) 5x^2+3x+5=0
x=-3+ ou -√9-100/10
x= -3+ou -√-91/10
não existe raízes nos reais
conjunto e vazio.

Answer 2

$\mathrm{\mathbf{a)}\ 3x^2-7x+4=0\ \to\ a=3\ \ \|\ \ b=-7\ \ \|\ \ c=4}\\\\ \mathrm{x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}=\dfrac{-(-7)\pm\sqrt{(-7)^2-4.3.4}}{2.3}=}\\\\ \mathrm{=\dfrac{7\pm\sqrt{49-48}}{6}=\dfrac{7\pm\sqrt{1}}{6}=\dfrac{7\pm1}{6}=\boxed{\dfrac{4}{3}}\ ou\ \boxed{1}}$

$\mathrm{\mathbf{b)}\ 9y^2-12y+4=0\ \to\ a=9\ \ \|\ b=-12\ \ \|\ \ c=4}\\\\ \mathrm{y=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}=\dfrac{-(-12)\pm\sqrt{(-12)^2-4.9.4}}{2.9}=}\\\\ \mathrm{=\dfrac{12\pm\sqrt{144-144}}{18}=\dfrac{12\pm0}{18}=\dfrac{12}{18}=\boxed{\dfrac{2}{3}}}$

$\mathrm{\mathbf{c)}\ 5x^2+3x+5=0\ \to\ a=5\ \ \|\ \ b=3\ \ \|\ \ c=5}\\\\ \mathrm{x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}=\dfrac{-3\pm\sqrt{3^2-4.5.5}}{2.5}=}\\\\ \mathrm{=\dfrac{-3\pm\sqrt{9-100}}{10}=\dfrac{-3\pm\sqrt{-91}}{10}=\dfrac{-3\pm\sqrt{91}\sqrt{-1}}{10}=}\\\\ \mathrm{=-\dfrac{3}{10}\pm\dfrac{\sqrt{91}}{10}i=\boxed{\mathrm{-\dfrac{3}{10}+\dfrac{\sqrt{91}}{10}i}}\ ou\ \boxed{\mathrm{-\dfrac{3}{10}-\dfrac{\sqrt{91}}{10}i}}}$