Question

ajuda ae nunca lhe pedir nada , vaie porfavor

Answer 1

Boa tarde Luana!

Solução!

$3(3y+1)=2(y^{2}-1)\\\\\\\ 9y+3=2y^{2} -2\\\\\\\ 2y^{2} -2-9y-3=0\\\\\\\\ 2y^{2} -9y-5=0 \Rightarrow a x^{2} +bx+c=0$

$y= \dfrac{-b\pm \sqrt{b^{2}-4.a.c } }{2.a}\\\\\\\ y= \dfrac{-(-9)\pm \sqrt{(-9)^{2}-4.2.(-5) } }{2.2}\\\\\\\ y= \dfrac{9\pm \sqrt{81+40 } }{4}\\\\\\\ y= \dfrac{9\pm \sqrt{121 } }{4}\\\\\\\ y= \dfrac{9\pm 11 }{4}\\\\\\ Raizes\\\\ y_{1}= \dfrac{9+11}{4}=\dfrac{20}{4}= 5\\\\\\ y_{2}= \dfrac{9-11}{4}= -\dfrac{2}{4}= -\dfrac{1}{2}$




$\boxed{Resposta:S=\{5,-\dfrac{1}{2} \}}$



Boa tarde!
Bons estudos!