Question

Quais as raízes da expressão

x . (4x-1) / 6 - (2-x)²  / 9 = 4x² + 3x +1 / 9

Answer 1

$\dfrac{x.(4x-1)}{6} - \dfrac{(2-x)^2}{9} = \dfrac{4x^2+3x+1}{9}\\\\ \dfrac{3x(4x-1) - 2(2-x)^2 = 2(4x^2+3x+1)}{\not{18}}\\\\ 3x(4x-1) - 2(2-x)^2 = 2(4x^2+3x+1)\\\\ 12x^2-3x - 2(4-4x+x^2) = 8x^2+6x+2\\\\ 12x^2-3x - 8+8x-2x^2 = 8x^2+6x+2\\\\ (12x^2 - 2x^2 -8x^2)+(-3x+8x-6x)+(-8-2) = 0\\\\ 2x^2-x-10 = 0\ \ (equa\c{c}\~ao\ do\ segundo\ grau)$



$Termos:\ \ \ a=2\ \ \ \ \ b=-1\ \ \ \ c=-10\\\\\\ Aplicando\ Bh\'askara:\\\\ x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}\\\\ x=\dfrac{1\pm\sqrt{1+80}}{4}\\\\ x=\dfrac{1\pm\sqrt{81}}{4}\\\\ x=\dfrac{1\pm9}{4}\\\\ Raizes:\\\\ x'=\dfrac{1+9}{4}\\\\ x'=\dfrac{10}{4}\\\\ \boxed{x'=\dfrac{5}{2}}\\\\\\ x''=\dfrac{1-9}{4}\\\\ x''=\dfrac{-8}{4}\\\\ \boxed{x''=-2}$

Bons estudos!