Question

determine o conjunto verdade das equações (x2-x-2).(-x2+2x+3) <0

Answer 1

$(2x - x - 2) \times 3 < 0 \\ (x - 2) \times 3 < 0 \\ 3x - 6 < 0 \\ 3x < 6 \\ x < 2$

Answer 2

Resposta:

x<2 e x>3

Explicação passo-a-passo:

$Aplicando~a~f\'{o}rmula~de~Bhaskara~para~x^{2}-x-2=0~~\\e~comparando~com~(a)x^{2}+(b)x+(c)=0,~temos~a=1{;}~b=-1~e~c=-2\\\\\Delta=(b)^{2}-4(a)(c)=(-1)^{2}-4(1)(-2)=1-(-8)=9\\\\x^{'}=\frac{-(b)-\sqrt{\Delta}}{2(a)}=\frac{-(-1)-\sqrt{9}}{2(1)}=\frac{1-3}{2}=\frac{-2}{2}=-1\\\\x^{''}=\frac{-(b)+\sqrt{\Delta}}{2(a)}=\frac{-(-1)+\sqrt{9}}{2(1)}=\frac{1+3}{2}=\frac{4}{2}=2\\\\S=\{-1,~2\}$

$Aplicando~a~f\'{o}rmula~de~Bhaskara~para~-x^{2}+2x+3=0~~\\e~comparando~com~(a)x^{2}+(b)x+(c)=0,~temos~a=-1{;}~b=2~e~c=3\\\\\Delta=(b)^{2}-4(a)(c)=(2)^{2}-4(-1)(3)=4-(-12)=16\\\\x^{'}=\frac{-(b)-\sqrt{\Delta}}{2(a)}=\frac{-(2)-\sqrt{16}}{2(-1)}=\frac{-2-4}{-2}=\frac{-6}{-2}=3\\\\x^{''}=\frac{-(b)+\sqrt{\Delta}}{2(a)}=\frac{-(2)+\sqrt{16}}{2(-1)}=\frac{-2+4}{-2}=\frac{2}{-2}=-1\\\\S=\{3,~-1\}$

Montando o comportamento dos sinais conforme gráfico e tabela.

temos x<2 e x>3