Question

3) Resolva as equações do 2º grau:

a) (x-3)² = -2x²

g) -2x² + 6x = 0

h) 2x² + x + 5 = 0

Answer 1

Resposta:

Explicação passo-a-passo:

$a)\\(x-3)^2=-2x^2\\\\x^2-6x+3^2=-2x^2\\\\3x^2-6x+9=0\ \ :3\\\\x^2-2x+3=0\\\\x_{1,\:2}=\frac{-\left(-2\right)\pm \sqrt{\left(-2\right)^2-4\cdot \:1\cdot \:3}}{2\cdot \:1}\\\\x_{1,\:2}=\frac{-\left(-2\right)\pm \:2\sqrt{2}i}{2\cdot \:1}\\\\x_1=\frac{-\left(-2\right)+2\sqrt{2}i}{2\cdot \:1}=\frac{2+2\sqrt{2}i}{2}=1+\sqrt{2}i \\\\\:x_2=\frac{-\left(-2\right)-2\sqrt{2}i}{2\cdot \:1}=\frac{2-2\sqrt{2}i}{2\cdot \:1}==1-\sqrt{2}i$

$b)\\-2x^2 + 6x = 0\\\\x(-2x+6)=0\ \ ,x_1=0\\\\-2x+6=0\\\\2x=6\\\\x=\frac{6}{2} \\\\x_2=3$

$c)\\2x^2 + x + 5 = 0\\\\x_{1,\:2}=\frac{-1\pm \sqrt{1^2-4\cdot \:2\cdot \:5}}{2\cdot \:2}=\frac{-1\pm \sqrt{39}i}{2\cdot \:2}\\\\x_1=\frac{-1+\sqrt{39}i}{4}\\\\\:x_2=\frac{-1-\sqrt{39}i}{4}$