Question

Olá boa noite vcs pode me ajuda com esse exercicio
Determine as raizes da funçaõ quadratica ,
a ) f ( x ) = x ² +2 x - 3

b) f ( x ) = x ²+ x - 12

Answer 1

a ) f ( x ) = x ² +2 x - 3

$\frac{d}{dx}\left(x^2+2x-3\right)\\\\=\frac{d}{dx}\left(x^2\right)+\frac{d}{dx}\left(2x\right)-\frac{d}{dx}\left(3\right)\\\\simplifico \\\\=2x+2-0\\\\=2x+2$

b) f ( x ) = x ²+ x - 12

$\frac{d}{dx}\left(x^2+x-12\right)\\\\=\frac{d}{dx}\left(x^2\right)+\frac{d}{dx}\left(x\right)-\frac{d}{dx}\left(12\right)\\\\simplifico\\\\=2x+1-0\\\\=2x+1$

Answer 2

a ) f ( x ) = x ² +2 x - 3

x²+2x-3=0

a=1

b=2

c=-3

formula do delta ∆= b²-4.a.c

substituindo..

∆= 2²-4.1.(-3)

∆= 4 + 12

∆= 16

Agora usando Bhaskara x= -b ± √∆/2.a

substituindo...

x= -2 ± √16/2.1

x= -2 ± 4/2

x'= -2 + 4/2 x"= -2 - 4/2

x'= 2/2 x"= -6/2

x'= 1 x"= -3

S={ 1,-3 }

b) f ( x ) = x ²+ x - 12

x²+x-12=0

a=1

b=1

c=-12

formula do delta ∆= b²-4.a.c

formula do delta ∆= b²-4.a.csubstituindo..

formula do delta ∆= b²-4.a.csubstituindo.. ∆= 1²-4.1.(-12)

)∆= 1 + 48

∆= 49

Agora usando Bhaskara x= -b ± √∆/2.a

Agora usando Bhaskara x= -b ± √∆/2.asubstituindo...

Agora usando Bhaskara x= -b ± √∆/2.asubstituindo...x= -1 ± √49/2.1

2.1x= -1 ± 7/2

/2x'= -1 + 7/2 x"= -1 - 7/2

/2x'= 6/2 x"= -8/2

2x'= 3 x"= -4

S={ 3,-4 }