Question

Calcule as raízes ou zeros da função quadratica abaixo:
-x²+2x-4=0

Answer 1

$x^2+2x-4=0\\ x_{1,\:2}=\frac{-b\pm \sqrt{b^2-4ac}}{2a}\\ \mathrm{Para\:}\quad a=-1,\:b=2,\:c=-4:\quad x_{1,\:2}=\frac{-2\pm \sqrt{2^2-4\left(-1\right)\left(-4\right)}}{2\left(-1\right)}\\ x=\frac{-2+\sqrt{2^2-4\left(-1\right)\left(-4\right)}}{2\left(-1\right)}:\quad 1-\sqrt{3}i\\ \frac{-2+\sqrt{2^2-4\left(-1\right)\left(-4\right)}}{2\left(-1\right)}\\ \mathrm{Remover\:os\:parenteses}:\quad \left(-a\right)=-a,\:-\left(-a\right)=a\\ =\frac{-2+\sqrt{2^2-4\cdot \:1\cdot \:4}}{-2\cdot \:1}$

$=\frac{-2+\sqrt{2^2-4\cdot \:1\cdot \:4}}{-2\cdot \:1}\\ =\frac{-2+\sqrt{12}i}{-2\cdot \:1}\\ \mathrm{Multiplicar\:os\:numeros:}\:2\cdot \:1=2\\ =\frac{-2+\sqrt{12}i}{-2}\\ \mathrm{Aplicar\:as\:propriedades\:das\:fracoes}:\quad \frac{a}{-b}=-\frac{a}{b}\\ =-\frac{-2+i\sqrt{12}}{2}\\ \sqrt{12}=2\sqrt{3}\\ =-\frac{-2+2\sqrt{3}i}{2}$

$\mathrm{Cancelar\:}\frac{-2+i2\sqrt{3}}{2}:\quad -1+\sqrt{3}i\\ =-\left(-1+\sqrt{3}i\right)\\ \mathrm{Colocar\:os\:parenteses}\\ =-\left(-1\right)-\left(\sqrt{3}i\right)\\ \mathrm{Aplicar\:as\:regras\:dos\:sinais}\\ -\left(-a\right)=a,\:\:\:-\left(a\right)=-a\\ =1-\sqrt{3}i$

$x=\frac{-2-\sqrt{2^2-4\left(-1\right)\left(-4\right)}}{2\left(-1\right)}:\quad 1+\sqrt{3}i\\ \frac{-2-\sqrt{2^2-4\left(-1\right)\left(-4\right)}}{2\left(-1\right)}\\ \mathrm{Remover\:os\:parenteses}:\quad \left(-a\right)=-a,\:-\left(-a\right)=a\\ =\frac{-2-\sqrt{2^2-4\cdot \:1\cdot \:4}}{-2\cdot \:1}\\ -2-\sqrt{2^2-4\cdot \:1\cdot \:4}=-2-\sqrt{12}i\\ =\frac{-2-\sqrt{12}i}{-2\cdot \:1}\\ \mathrm{Multiplicar\:os\:numeros:}\:2\cdot \:1=2\\ =\frac{-2-\sqrt{12}i}{-2}$

$\mathrm{Aplicar\:as\:propriedades\:das\:fracoes}:\quad \frac{a}{-b}=-\frac{a}{b}\\ =-\frac{-2-i\sqrt{12}}{2}\\ \sqrt{12}=2\sqrt{3}\\ =-\frac{-2-2\sqrt{3}i}{2}\\ \mathrm{Cancelar\:}\frac{-2-i2\sqrt{3}}{2}:\quad -1-\sqrt{3}i\\ =-\left(-1-\sqrt{3}i\right)\\ \mathrm{Colocar\:os\:parenteses}\\ =-\left(-1\right)-\left(-\\ \sqrt{3}i\right)\\ \mathrm{Aplicar\:as\:regras\:dos\:sinais}\\ -\left(-a\right)=a$

$=1+\sqrt{3}i\\ \mathrm{As\:solucoes\:para\:a\:equacao\:de\:segundo\:grau\:sao:\:}\\ x=1-\sqrt{3}i,\:x=1+\sqrt{3}i$

Bons Estudos!!