Question

Ache as raízes das equações:
a) x² - 4x - 5 =
b) x² - 14x + 48 = 0

Answer 1

a
x² - 4x - 5 = 0
delta = 16 + 20 = 36 ou +-V36 = +-6 ***

x = ( 4 +-6)/2
x1 = 10/2 = 5***
x2 = -2/2 = -1 ***

b
x² - 14x  + 48 = 0
delta = 196 - 192  = 4 ou +-V4 = +-2 ****

x = ( 14 +-2)/2
x1 = 16/2 = 8 ***
x2 = 12/2 = 6 ***

Answer 2

Olá.

a) x² - 4x - 5 = 0

$\mathsf{x^{2} - 4x - 5 = 0} \\ \\ \\ \mathsf{\triangle = b^{2} - 4ac} \\ \\ \mathsf{\triangle = 16 - 4 \times 1 \times (-5)} \\ \\ \mathsf{\triangle = 36} \\ \\ \\ \mathsf{x = \dfrac{-b \pm \sqrt{\triangle}}{2a}} \\ \\ \\ \mathsf{x_{1} = \dfrac{4 + 6}{2} = 5} \\ \\ \\ \mathsf{x_{2} = \dfrac{4 - 6}{2} = -1}$

$\mathsf{\boxed{S = {\{-1, \: 5\}}}}$

b) x² - 14x + 48 = 0

$\mathsf{x^{2} - 14x + 48 = 0} \\ \\ \\ \mathsf{\triangle = b^{2} - 4ac} \\ \\ \mathsf{\triangle = 196 - 4 \times 1 \times 48} \\ \\ \mathsf{\triangle = 4} \\ \\ \\ \mathsf{x = \dfrac{-b \pm \sqrt{\triangle}}{2a}} \\ \\ \\ \mathsf{x_{1} = \dfrac{14 + 2}{2} = 8} \\ \\ \\ \mathsf{x_{2} = \dfrac{14 - 2}{2} = 6}$

$\mathsf{\boxed{S = {\{6, \: 8\}}}}$

Bons estudos.