Question

como se resolve essas equações do segundo grau utilizando a formula de BHASKARA : x²+6x+9=0 e x²-2x-8=0 ????

Answer 1

x² + 6x + 9 = 0 
x = -b $+- \sqrt{x} b² - 4 . a . c / 2$
x = -6 +- $\sqrt{x} 36 - 36$ / 2
x = -6 +- 0 / 2
x= -6 / 2
x = -3

x² - 2x - 8 = 0
x = 2 +- $\sqrt{x} 4 + 32$ / 2 
x = 2 +- $\sqrt{x} 36$ / 2 
x = 2 +- 6 / 2
x' = 2 + 6 / 2 = 8 / 2 = 4
x'' = 2 - 6 /2 = -4 / 2 = -2

Answer 2

: x²+6x+9=0
a=1 b=6 c=9

∆=b*2-4 a.c
∆=6*2-4.1.9
∆=36-36
∆= 0

-b+- √∆/2.a
-6+-√0/2.1
-6+0/2 = -6/2 = -3
-6-0/2 = -6/2 = -3

x*2-2x-8=0
a=1 b= -2 c= -8
b*2-4.a.c
(-2)*2-4.1.(-8)
4+32
∆=36

-b+-√∆/2.a
-(-2)+-√36/2.1
4+-6/2
4+6/2= 10/2= 5
4-6/2= -2/2= -1