Question

a= -3/2, b= 1/3 e c = 1/3
me ajude .. responda aí precisa dessa resposta

Answer 1

Kassandra

Nada difícil. Só substituir os valores e efetuar operações indicadas
Veja
                 $b^2 - 4.a.c \\ \\ =( \frac{1}{3})^2 - 4.( -\frac{3}{2}).( \frac{1}{3)} \\ \\ = \frac{1}{9} +2 \\ \\ = \frac{1}{9}+ \frac{18}{9}$

                         $=\frac{19}{9}$  RESULTADO FINAL

As raízes são determinadas pela fórmula resolutiva (Bháskara)

                       $x= \frac{-b+/- \sqrt{x} }{y} \\ \\ = \frac{ -\frac{1}{3}+/- \frac{ \sqrt{19} }{9} }{2(- \frac{3}{2}) } \\ \\ = \frac{ \frac{-1+/- \sqrt{1} }{3} }{-3} \\ \\ = \frac{-1+/- \sqrt{19} }{-9} \\ \\ = \frac{1-/+ \sqrt{19} }{9} \\ \\ x1= \frac{1- \sqrt{19} }{9} \\ x2= \frac{1+ \sqrt{19} }{9}$

                               S = {$\frac{1- \sqrt{19} }{9}, \frac{1+ \sqrt{19} }{9}$}
                  

Answer 2

Vamos lá ! 

$delta = b^{2} -4.a.c\\\\\\delta =( \frac{1}{3})^{2} -4.(- \frac{3}{2} ).( \frac{1}{3} )\\\\\\delta= \frac{1}{9} -4.( -\frac{3}{6} )\\\\\\delta= \frac{1}{9} + \frac{12}{6} \\\\\\delta= \frac{19}{9}$

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Encontrei delta , agora quero as raízes : 

$x= \frac{-b+- \sqrt{delta} }{2a} \\\\\\x= \frac{- \frac{1}{3}+- \sqrt{ \frac{19}{9} } }{2. -\frac{3}{2} } \\\\\\x= \frac{ -\frac{1}{3} +- \frac{ \sqrt{19} }{3} }{ -\frac{6}{2} } \\\\\\x= \frac{- \frac{1}{3}+- \frac{ \sqrt{19} }{3} }{- 3} \\\\\\x'= ( -\frac{1}{3} + \frac{ \sqrt{19} }{3} }).{- \frac{1}{3} } \\\\\\x'= \frac{1}{9} }{- \frac{ \sqrt{19} }{9} } \\\\\\\boxed{x'= \frac{1- \sqrt{19}}{9} }$

$x''= \frac{- \frac{1}{3}- \frac{ \sqrt{19} }{3} }{- \frac{3}{1} } \\\\\\x''= ( -\frac{1}{3} - \frac{ \sqrt{19} }{3} }).{ -\frac{1}{3} } \\\\\\x''= \frac{1}{9} +\frac{ \sqrt{19} }{9} \\\\\\\boxed{x''= \frac{1+ \sqrt{19} }{9} }$
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Então temos como solução : 


$\boxed{\boxed{S= \{\ \frac{1- \sqrt {19}}{9} \ ,\ \frac{1+ \sqrt{19 }}{9} \ \ \}}}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ ok$