Question

Resolva as seguintes equações do 2° grau, identifique os coeficientes e determine a raiz se existir​

Answer 1

Resposta:

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}