Question

Utilize a fórmula de Bharkara e ​resolva:
a)
$6x - 4 {x}^{2} + 10 = 0$
b)
$4x - 8 + 2 {x}^{2} - 6 = 0$
c)
$10 - {x}^{2} - 6 {x}^{2} = 0$
d)
$4{x}^{2} - 8x = 0$
e)
$12 - {x}^{2} = 0$




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Answer 1

Em bhaskara, temos:

Delta = b² -4.a.c

X = -b+-Vdelta / 2.A

Nas questões:

a) A = -4

B = 6  

C = 10

Delta = 36 -4.-4.10

Delta = 36+160

Delta = 196

X = -6 +-V196 / -8

X1 = -6 + 14 / -8

X1 = 8 / -8

X1 = -1

X2 = -6 – 14 / -8

X2 = -20 / -8

X2 = 2,5 ou 5/2

B) A = 2

B = 4  

C = -14

Delta = 16 – 4.2.-14

Delta = 16 + 112

Delta = 128

X = -4 +-V128 / 4

X = -1 +-V128

X1 = -1 + V128 = -1+8V2 ou -1+11,3 = 10,3

X2 = -1 – V128 = -1-8V2 ou -1-11,3 = -12,3

C) A = -7

B = 0  

C = 10

Delta = 0² - 4.-7.10

Delta = +280

X = +-V280 / -14

X1 = +16,7 / -14  = -1,20

X2 = -16,7 / -14 = 1,20

D)A = 4

B =-8

C= 0

Delta = 64 -0

Delta = 64

X = 8 +-8 / 8

X1 = 8+8/8 = 16/8 = 2

X2 = 8-8 / 8 = 0/8 = 0

E) A = -1

B = 0

C = 12

Delta = 0 +4.1.12

Delta = 48

X = +-V48 / -2

X1 = +6,9 / -2 = -3,45

X2 = -6,9 / -2 = 3,45