Question

a)x²+4x+1=0 b)x²-x+5=0 c)x²-5x-6=0 d) x²=0 e)x²+10x+9=0 f)y²+8y+16=0.

Answer 1

Resposta:

Explicação passo-a-passo:

a)x²+4x+1=0

Δ= b² - 4.a.c

Δ= 4² - 4.1.1

Δ= 16 - 4

Δ= 12

x= -4±$\sqrt{12} \sqr$/ 2.1

x= -4 ±2$\sqrt{3}$ /2

x'= -4 - 2$\sqrt{3}$ /2

x'=  -2 -$\sqrt{3}$

x" = -4+2$\sqrt{3}$ /2

x" = -2+$\sqrt{3}$

b)x²-x+5=0

Δ= b² - 4.a.c

Δ= (-1)² - 4.1.5

Δ= 1-20

Δ= -19

Não existe raízes reais

c)x²-5x-6=0

Δ= b² - 4.a.c

Δ=( -5)² - 4.1.(-6)

Δ= 25+  24

Δ= 49

x= -(-5) ±$\sqrt{49}$/2

x=5±7/2

x'= 5-7/2

x' = -2/2

x' = -1

x"= 5+7/2

x"= 12/2

x"= 6

s = (-1 ,6)

d)x²=0

x²=$\sqrt{0}$

x= 0

e)x²+10x+9=0

Δ= b² - 4.a.c

Δ=( 10)² - 4.1.(9)

Δ= 10 -  36

Δ= 64

x= -(10) ±$\sqrt{64}$/2

x= -10±8/2

x'= -10+8/2

x' = -2/2

x' = -1

x"= -10-8/2

x"=  -18/2

x"= - 9

s = (-9 ,-1)

f)y²+8y+16=

Δ= b² - 4.a.c

Δ=( 8)² - 4.1.(16)

Δ= 64 -  64

Δ= 0

y= -(8) ±$\sqrt{0}$/2

y'= -8 ± 0/2

 y'= 8+ 0/2

y'= -8/2

y' = -4

y"= -8-0/2

y"=  -8/2

y"= - 4

s = (-4)  Existe apenas uma raiz real