Question

1-resolva as equacoes qudraticas usando a formula de Bhaskara;
a) x²-6x =0
b)x²+x+2=0
c)x²-12x + 35=0
d)-x²-15x-54=0
e)x²-2x-63=0
f)x²-4 =0
g)x²-64=0
h)x²- 11x+28=0

resumido pfvr gnt;)

Answer 1

Resposta:

ESPERO AJUDAR...

Explicação passo-a-passo:

a) x²-6x =0

Δ=-6²- 4.1.0

Δ= 36 -0 = 36

X= -(-6) +- √ 36 / 2.1

X= 6 +-6 / 2

X'= 6 +6 /2= 6  

X"= 6-6/2 = 0

b)x²+x+2=0

Δ=1²- 4.1.2

Δ= 1-8 = -7

c)x²-12x + 35=0

Δ=(-12)²- 4.1.35

Δ= 144 - 140 = 4

X= -(-12) +- √ 4 / 2.1

X= 12 +-2 / 2

X'= 12 +2 /2= 7  

X"= 12-2/2 = 5

d)-x²-15x-54=0

Δ=(-15)²- 4.(-1).(-54)

Δ= 225-216 = 9

X= -(-15) +- √ 9 / 2.(-1)

X= 15 +-3 / -2

X'= 15 +3 /-2 = -9

X"= 15-3/-2 = -6

e)x²-2x-63=0

Δ=(-2)²- 4.1.63

Δ= 4-252 = 256

X= -(-2) +- √ 256 / 2.1

X= 2 +-16 / 2

X'= 2 +16 /2= 9  

X"= 2-16/2 = -7

f)x²-4 =0

Δ=0²- 4.1.(-4)

Δ= 0+16 = 16

X= 0 +- √ 16 / 2.1

X= 0 +-4 / 2

X'= 0 +4 /2= 2  

X"= 0-4/2 = -2

g)x²-64=0

Δ=0²- 4.1.(-64)

Δ= 0+256 = 256

X= 0 +- √ 256 / 2.1

X= 0 +-16 / 2

X'= 0 +16 /2= 7  

X"= 0-16/2 = -7

h)x²- 11x+28=0

Δ=(-11)²- 4.1.28

Δ= 121-112 = 9

X= -(-11) +- √ 9 / 2.1

X= 11 +-3 / 2

X'= 11 +3 /2= 7  

X"= 11-3/2 = 4