Question

Mi ajudem nessa questão preciso mt

Answer 1

a) 4x² - 36 = 0

4x² = 36

x² = 36/4

x² = 9

x = ±$\sqrt{9}$

x' = 3  |  x'' = -3

b) x² + 9 = 0

x² = -9

x = ±$\sqrt{-9}$ (Inexistente)

c) x² - 6x + 9 = 4

x² - 6x + 5 = 0

Δ= (-6)² - 4 . 1 . 5

Δ= 36 - 20

Δ= 16

x= 6±√16 / 2.1

x' = $\frac{6+4}{2}$

x' = 5

x'' = $\frac{6-4}{2}$

x'' = 1

d) x² + 10x + 25 = 0

Δ= (10)² - 4 . 1 . 25

Δ= 100 - 100

Δ= 0

x = $\frac{-10}{2}$

x = -5

e) x (4x + 4) = 0

x' = 0

4x + 4 = 0

4x = -4

x= -4/4

x''= -1

f) x (-5x + 10) = 0

x' = 0

-5x + 10 = 0

-5x = -10 (-)

5x = 10

x = 10/5

x'' = 2