Question

ajudem pff,
a ) -x² +x+12=0
b)-x² +6x -5=0
c)6x² +x-1= 0
d)3x²-7x+2=0
e)2x²-7x=15
f)4x²+9=12x

Answer 1

a) -x² +x +12 = 0
∆= 1² -4.(-1).12
∆= 1 + 48
∆= 49

x= -1 ± 7/ -2
x' = -3
x" = 4

b) -x² +6x -5 = 0
∆= 36 -4.(-1).(-5)
∆= 36 - 20
∆= 16

x = -6 ± 4/-2
x'= 1
x" = 5

c)6x² +x -1 = 0
∆= 1² -4.6.(-1)
∆= 1 +24
∆= 25

x = -1 ±5/12
x'= 1/3
x"= -1/2

d) 3x² -7x +2 = 0
∆= 49 -4.3.2
∆= 49 - 24
∆= 25

x = 7 ±5/6
x' = 12/6 = 2
x" = 1/3

e)2x² -7x = 15
2x² -7x -15 = 0
∆= 49 -4.2.(-15)
∆= 49 + 120
∆= 169

x = 7 ±13/4
x' = 20/4 = 5
x" = -3/2

f) 4x² +9= 12x
4x² +9 -12x = 0
∆= 12² -4.4.9
∆= 144 - 144
∆= 0

x = 12 ± 0/8
x' = 12/8 = 3/2
x" = 12/8 = 3/2

Espero ter ajudado!!