Question

Qual a solução de x²+6x+9=49 e de x²+2x+1=100

Answer 1

x²+6x+9=49
x²+6x-40=0

delta= 6²-4.1.(-40)
delta= 36+160
delta= 196

x= -6±√196/2.1

x= -6±14/2

x= -6+14/2= 8/2= 4

x= -6-14/2= -20/2= -10

s={ 4,-10 }



x²+2x+1=100
x²+2x-99=0

delta= 2²-4.1.(-99)
delta= 4+396
delta= 200

x= -2±√200/2.1

x= -2±20/2

x= -2+20/2= 18/2= 9

x= -2-20/2= -22/2= -11

s={ 9,-11 }

Answer 2

+/- : (mais ou menos)
$x^{2} +6x+9=49 \\ \\ x^{2} +6x+9-49=0 \\ \\ x^{2} +6x-40=0$
Δ = b² - 4ac
Δ = 6² - 4 . 1 . (-40)
Δ = 36 + 160
Δ = 196

x = (-b +/- √Δ)/2a
x' = (-6 + √196)/2 . 1
x' = (-6 + 14)/2
x' = 8/2
x' = 4

x" = (-6-14)/2
x" = -20/2
x" = -10

-----------------------------------

$x^{2} +2x+1=100 \\ \\ x^{2} +2x+1-100=0 \\ \\ x^{2} +2x-99=0$

Δ = b² - 4ac
Δ = 2² - 4 . 1 . (-99)
Δ = 4 + 396
Δ = 400

x = (-b +/- √Δ)/2a
x' = (-2 + √400)/2 . 1
x' = (-2 + 20)/2
x' = 18/2
x' = 9

x" = (-2-20)/2
x" = -22/2
x" = -11