Question

Obtenha a solução da equação modular |x2-3x|=4​

Answer 1

Resposta:

$S=\{-1,~4,~\frac{3-i\sqrt{7}}{2},~\frac{3+i\sqrt{7}}{2}\}$

Explicação passo-a-passo:

|x²-3x|=4

x²-3x=4 => x²-3x-4=0   (I)

ou

-x²+3x=4 => x²-3x+4=0  (II)

Resolvendo as equações:

(I)

x²-3x-4=0

$Aplicando~a~f\'{o}rmula~de~Bhaskara~para~x^{2}-3x-4=0~~\\e~comparando~com~(a)x^{2}+(b)x+(c)=0,~temos~a=1{;}~b=-3~e~c=-4\\\\\Delta=(b)^{2}-4(a)(c)=(-3)^{2}-4(1)(-4)=9-(-16)=25\\\\x^{'}=\frac{-(b)-\sqrt{\Delta}}{2(a)}=\frac{-(-3)-\sqrt{25}}{2(1)}=\frac{3-5}{2}=\frac{-2}{2}=-1\\\\x^{''}=\frac{-(b)+\sqrt{\Delta}}{2(a)}=\frac{-(-3)+\sqrt{25}}{2(1)}=\frac{3+5}{2}=\frac{8}{2}=4\\\\S=\{-1,~4\}$

(II)

x²-3x+4=0

$Aplicando~a~f\'{o}rmula~de~Bhaskara~para~x^{2}-3x+4=0~~\\e~comparando~com~(a)x^{2}+(b)x+(c)=0,~temos~a=1{;}~b=-3~e~c=4\\\\\Delta=(b)^{2}-4(a)(c)=(-3)^{2}-4(1)(4)=9-(16)=-7\\\\\sqrt{\Delta}=\sqrt{-7} =\sqrt{-1} .\sqrt{7} =i\sqrt{7} \\\\x^{'}=\frac{-(b)-\sqrt{\Delta}}{2(a)}=\frac{-(-3)-i\sqrt{7}}{2(1)}=\frac{3-i\sqrt{7}}{2}\\\\x^{''}=\frac{-(b)+\sqrt{\Delta}}{2(a)}=\frac{-(-3)+i\sqrt{7}}{2(1)}=\frac{3+i\sqrt{7}}{2}\\\\S=\{\frac{3-i\sqrt{7}}{2},~\frac{3+i\sqrt{7}}{2}\}$