Question

6. resolva, em R, as equações a seguir: A) (-x²+1)•(x²-3x+2)=0​

Answer 1

Resposta:

S={-1,1,2}, raiz dupla em 1

Explicação passo-a-passo:

(-x²+1)•(x²-3x+2)=0

(-x²+1)=0 (I)

ou

(x²-3x+2)=0 (II)

(I)

(-x²+1)=0

-x²+1=0

x²=1

x=±√1=±1

de (II)

(x²-3x+2)=0

x²-3x+2=0

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