Question

dertemine o as raizes da equação de 2° grau:
x²-2x-3​

Answer 1

x² - 2x - 3​ = 0

a = 1

b = -2

c = -3

$\Delta = b^{2} -4.a.c$

Δ = (-2)² - 4 . 1 . (-3)

Δ = 4 + 12

Δ = 16

$x = \dfrac{-b \pm \sqrt{\ \Delta}}{2.a}$

$x = \dfrac{-(-2) \pm \sqrt{16}}{2.1}$

$x = \dfrac{2 \pm {4}}{2}$

$x_1 = \dfrac{2 - \qrt{4}}{2} = \dfrac{-2}{2} = -1$

$x_2 = \dfrac{2  +{4}}{2} = \dfrac{6}{2} =3$

Answer 2

Resposta:

$\boxed{\mathtt{S = \left \{ - 1, 3 \right \}}}$

Explicação passo-a-passo:

$\\ \displaystyle \mathsf{x^2 - 2x - 3 = 0} \\\\ \mathsf{x^2 + (x - 3x) - 3 = 0} \\\\ \mathsf{x^2 + x - 3x - 3 = 0} \\\\ \mathsf{x(x + 1) - 3(x + 1) = 0} \\\\ \mathsf{(x + 1) \cdot \left [ x - 3 \right ] = 0} \\\\ \boxed{\boxed{\mathsf{S = \left \{ - 1, 3 \right \}}}}$