Question

Identifique os coeficientes :
a)) x2 - 5x + 6 = 0
b) 2x2 - 8x + 8 = 0
c) x2 - 4x - 5 = 0
d) -x2 + x + 12 = 0
e) 6x2 + x - 1 = 0
f) 3x2 - 7x + 2 = 0
g) 2x2 - 7x = 15
h) 4x2 + 9 = 12x
i) x2 = x + 12
j) 2x2 = -12x - 18
k) x2 + 9 = 4x
l) 25x2 = 20x – 4

Answer 1

Resposta:

A- x² - 5x + 6 = 0

a = 1

b = -5

c = 6

Delta:

Δ = b² - 4ac

Δ = (-5)² - 4 * 1 * 6

Δ = 25 - 24

Δ = 1

    Bhaskara:

    x = - b ± √Δ / 2a

    x = - (-5) ± √1 / 2 * 1

    x = 5 ± 1 / 2

    x' = 5 + 1 / 2 = 6 / 2 = 3

    x'' = 5 - 1 / 2 = 4 / 2 = 2

S = {2, 3}

b- 2x2 - 8x + 8 = 0

2 ( 2 − 4 + 4 ) = 0

S= 2

C-  x2 - 4x - 5 = 0

− 2+ + 1 2 = 0

− ( 2 − − 1 2 ) = 0

x=5

x=-1

D- -x2 + x + 12 = 0

− 2 + + 1 2 = 0

− ( 2 − − 1 2 ) = 0

X=4

X=-3

E-  6x2 + x - 1 = 0

− ± 2 − 4 √ 2

Solução

= 1 /3

= − 1/ 2

F- 3x2 - 7x + 2 = 0

3 2 − 7 + 2 = 0

= 3

 

= − 7

= 2

= − ( − 7 ) ± ( − 7 ) 2 − 4 ⋅ 3 ⋅ 2 √ 2 ⋅ 3

Solução

= 2

= 1/ 3

G-3x2 - 7x + 2 = 0

3 2 − 7 + 2 = 0

= 3

= − 7

b=−7 = 2 c=2 = − ( − 7 ) ± ( − 7 ) 2 − 4 ⋅ 3 ⋅ 2 √ 2 ⋅ 3

Solução

= 2

= 1/ 3

H-  4x2 + 9 = 12x

4 2 +9 = 1 2

4 2 + 9 − ( 1 2 ) = 0

Solução

= 3/ 2

I- x2 = x + 12

2x=-x +12

12/3

x = 4

j-  2x2 = -12x - 18

2x + 12X  = -18- 2

X=20/10

X= 0.5

K-  x2 + 9 = 4x

2X - 4X= -9

X=9/2

X=4.5

L-  25x2 = 20x – 4

25x² = 20x -4

25x² -20x +4

Δ= b²-4ac

Δ= (-20)² -4.25.4

Δ= 400 -400

Δ= 0

x= 2/5

ESPERO TER AJUDADO