Aguem sabe me reponder
x²+2x+1=81
Soluções para a tarefa
Respondido por
11
x²+2x+1=81
x²+2x-80=0
Δ=b²-4ac
Δ=2²-4×1×(-80)
Δ=4+320
Δ=324
x'=-b+√Δ÷2×a
x'=-2+18÷2×1
x'=16÷2
x'=8
x"=-2-18÷2
x"=-20÷2
x"=-10
S={8,-10}
x²+2x-80=0
Δ=b²-4ac
Δ=2²-4×1×(-80)
Δ=4+320
Δ=324
x'=-b+√Δ÷2×a
x'=-2+18÷2×1
x'=16÷2
x'=8
x"=-2-18÷2
x"=-20÷2
x"=-10
S={8,-10}
Respondido por
3
x²+2x+1=81
x² + 2x + 1 - 81 = 0
x² + 2x - 80 = 0
a = 1 b = + 2 c = - 80
Δ = b² - 4.a.c
Δ = (2)² - 4.(1).(-80)
Δ = 4 + 320
Δ = 324
x = - b ± √Δ
2.a
x = - (+2) ± √324
2.1
x = - 2 ± 18
2
x' = - 2 + 18 = 16 = 8
2 2
x"= - 2 - 18 = - 20 = - 10
2 2
S[- 10 ; 8]
x² + 2x + 1 - 81 = 0
x² + 2x - 80 = 0
a = 1 b = + 2 c = - 80
Δ = b² - 4.a.c
Δ = (2)² - 4.(1).(-80)
Δ = 4 + 320
Δ = 324
x = - b ± √Δ
2.a
x = - (+2) ± √324
2.1
x = - 2 ± 18
2
x' = - 2 + 18 = 16 = 8
2 2
x"= - 2 - 18 = - 20 = - 10
2 2
S[- 10 ; 8]
Perguntas interessantes