Question

(x+4)²=9x+22   Resolva em formula de bhaskara
Ajudem,por favor!

Answer 1

(x+4)² = 9x+22
(x+4) x (x+4) = 9x+22
x²+4x+4x+16 = 9x+22
x²+8x+16 = 9x+22
x²+8x+16-9x-22 = 0
x²-x-6 = 0
a= 1   b =-1   c= -6
Delta = b²-4xaxc
Delta = (-1)²-4x1x(-6)
Delta = 1+24
Delta =25

x= -b+-$\sqrt{delta}$/2xa
x= -(-1)+-$\sqrt{25}$/2x1
x= 1+-5/2
x'= 1+5/2    x'= 6/2    x'=3
x"= 1-5/2    x" = -4/2  x"= -2

Answer 2

(x + 4)^2 = 9x + 22
Desenvolvendo (x + 4)^2
x^2 + 2.x.4 + 4^2 = 9x + 22
x^2 + 8x + 16 = 9x + 22
x^2 +8x - 9x + 16 - 22 = 0
x^2 - x - 6 = 0
∆ = (-1)^2 - 4.1.(-6)
∆ = 1 + 24
∆ = 25

x' = {-(-1) + √25}/2.1
x' = (1 + 5)/2
x' = 6/2
x' = 3

x" = {-(-1) - √25}/2.1
x" = (1 - 5)/2
x" = -4/2
x" = -2

S = {3, -2}