Question

resolva as seguintes equações do 2º grau , identifique os coeficientes e determine as raizes se existir.

a) x²-5+6=0

b) x²+2x-8=0

c) x²-5x+8=0

d)-x²+x+12=0

e)-x²+6x-5=0

f)6x²+x-1=0

h) 2x²-7x+2=0

i) 4x²+9=12x

j) 2x²= -12x-18

Answer 1

a) x²-5+6=0
x² +1 = 0

a=1   b= 0      c= +1
▲= b² -4.a.c
▲=(0)² -4.(1).(+1)
▲= 0 -4
▲= -4  não existe raiz real

b) x²+2x-8=0

a= 1     b= +2      c= -8
▲= b² -4.a.c
▲=(2)² -4.(1).(-8)
▲=4 +32
▲= 36

x= (-b ± √▲)/2.a

x= [-(+2)  ± √36]/2.1

x= [-2  ± 6]/2

x'=[-2 +6]/2 = +4/2 = 2

x"=[-2-6]/2 = -8/2 = -4

S{2 ; -4}


c) x²-5x+8=0

a= 1      b= -5       c= +8
▲= b² -4.a.c
▲=(-5)² -4.(1).(+8)
▲=25-32
▲= -7  não existe raiz real.


d)-x²+x+12=0.(-1)
x²-x-12=0

a= 1        b= -1         c= -12
 ▲= b² -4.a.c
▲=(-1)² -4.(1).(-12)
▲=1 +48
▲= 49

x= (-b ± √▲)/2.a

x= [-(-1)  ± √49]/2.1

x= [+1  ±7]/2

x'=[+1 +7]/2 = +8/2 = 4

x"=[1-7]/2 = -6/2 = -3

S{4 ; -3}

e)-x²+6x-5=0.(-1)
x²-6x+5=0

a= 1    b= -6      c= +5
▲= b² -4.a.c
▲=(-6)² -4.(1).(+5)
▲=36 -20
▲= 16

x= (-b ± √▲)/2.a

x= [-(-6)  ± √16]/2.1

x= [+6  ±4]/2

x'=[+6+4]/2 = +10/2 = 5

x"=[6-4]/2 =2/2 = 1

S{5 ; 1}


f)6x²+x-1=0

a= 6        b= +1        c= -1
▲= b² -4.a.c
▲=(1)² -4.(6).(-1)
▲=1 +24
▲= 25

x= (-b ± √▲)/2.a

x= [-(+1)  ± √25]/2.6

x= [-1 ±5]/12

x'=[-1 +5]/12 = 4/12      simplifica 4 e 12 por 4 = 1/3

x"=[-1 -5 ]/2 = - 6/12  simplifica -6 e 12 por 6 = -1/2

S{1/3; -1/2}

h) 2x²-7x+2=0

a=  2       b= -7         c= +2
▲= b² -4.a.c
▲=(-7)² -4.(2).(+2)
▲=49 - 16
▲= 33

x= (-b ± √▲)/2.a

x= [-(-7)  ± √33]/2.2

x= [+7± √33 ]/4

x'=[7+ √33]/4 

x"=[7-√33]/4 

S[7+ √33]/4  ; [7- √33]/4 


i) 4x²+9=12x
 4x²-12x +9=0

a= 4        b= -12        c= +9
▲= b² -4.a.c
▲=(-12)² -4.(4).(+9)
▲=144 - 144
▲= 0

x= (-b ± √▲)/2.a

x= [-(-12)  ± √0]/2.4

x= [+12± 0 ]/8

x'=[12+ 0]/8 = 12/8 simplificando 12 e 8 por 4= 3/2

x"=[12-0]/8  = 12/8 = 3/2

S{3/2}


j) 2x²= -12x-18
 2x²+12x+18 =0

a= 2     b= +12      c= +18
▲= b² -4.a.c
▲=(12)² -4.(2).(+18)
▲=144 - 144
▲=0

x= (-b ± √▲)/2.a

x= [-(+12)  ± √0]/2.2

x= [-12± √0 ]/4

x'=[-12+ 0]/4  = -12/4 = -3

x"=[-12-0]/4 = -12/4 = -3

S{-3}