Question

Quais são as raízes da equações abaixo



2
X +4x +4=0

2
X -X -6=0

2
X +8x +12=0

Answer 1

Resposta:

1) x₁=x₂ = -2

2) x₁= 3 e x₂= -2

3) x₁= -6 e x₂= -2

Explicação passo-a-passo:

Para resolver os exercícios vamos utilizar a fórmula resolutiva:

1) x² + 4x + 4 = 0

  a = 1; b = 4; c = 4

$x=\dfrac{-b\pm\sqrt{b^2 - 4\times a\times c}}{2\times a}\\\\\\x=\dfrac{-4\pm\sqrt{4^2 - 4\times 1\times 4}}{2\times 1}=\dfrac{-4\pm\sqrt{16 -16}}{2}=\dfrac{-4\pm\sqrt{0}}{2}\\\\x_1=\dfrac{-4-\sqrt{0}}{2}=\dfrac{-4}{\ \ 2}=-2\\\\x_2=\dfrac{-4+\sqrt{0}}{2}=\dfrac{-4}{\ \ 2}=-2\\\\\boxed{\bf{x_1=x_2=-2}}$

2) x² - x - 6 = 0

   a = 1; b = -1; c = -6

$x=\dfrac{-b\pm\sqrt{b^2 - 4\times a\times c}}{2\times a}\\\\\\x=\dfrac{-(-1)\pm\sqrt{(-1)^2 - 4\times 1\times (-6)}}{2\times 1}=\dfrac{+1\pm\sqrt{1 +24}}{2}=\dfrac{1\pm\sqrt{25}}{2}=\dfrac{1\pm5}{2}\\\\x_1=\dfrac{1+5}{2}=\dfrac{6}{2}=3\\\\x_2=\dfrac{1-5}{2}=\dfrac{-4}{\ \ 2}=-2\\\\\boxed{\bf{x_1=3}} \ e\ \boxed{\bf{x_2=-2}}$

3) x² + 8x + 12 = 0

   a = 1; b = 8; c = 12

$x=\dfrac{-b\pm\sqrt{b^2 - 4\times a\times c}}{2\times a}\\\\\\x=\dfrac{-8\pm\sqrt{8^2 - 4\times 1\times 12}}{2\times 1}=\dfrac{-8\pm\sqrt{64 -48}}{2}=\dfrac{-8\pm\sqrt{16}}{2}=\dfrac{-8\pm4}{2}\\\\x_1=\dfrac{-8-4}{2}=\dfrac{-12}{\ \ \ \ 2}=-6\\\\x_2=\dfrac{-8+4}{2}=\dfrac{-4}{\ \ 2}=-2\\\\\boxed{\bf{x_1=-6}}\ e\ \boxed{\bf{x_2=-2}}$

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