Determine as soluções reais das equações :
A) (x+1)²=3+x
B) (t-1)²+(t+2)²-9=0
Soluções para a tarefa
Respondido por
0
a)
(x + 1)² = 3 + x
x² + 2x + 1 = 3 + x
x² + 2x - x + 1 - 3 = 0
x² + x - 2 = 0
a = 1; b = 1; c = - 2
Δ = b² - 4ac
Δ = 1² - 4.1.(-2)
Δ = 1 + 8
Δ = 9
x = - b +/- √Δ = - 1 +/- √9
2a 2.1
x = - 1 + 3 = 2/2 = 1
2
x = - 1 - 3 = - 4/2 = - 2
2
R.: x = 1 e x = - 2
--------------------------------------------------
b)
(t - 1)² + (t + 2)² - 9 = 0
t² - 2t + 1 + t² + 2.2t + 4 - 9 = 0
t² - 2t + 1 + t² + 4t - 5 = 0
t² + t² - 2t + 4t + 1 - 5 = 0
2t² + 2t - 4 = 0 (:2)
t² + t - 2 = 0
a = 1; b = 1; c = - 2
Δ = b² - 4ac
Δ = 1² - 4.1.(-2)
Δ = 1 + 8
Δ = 9
x = - b +/- √Δ = - 1 +/- √9
2a 2.1
x = - 1 + 3 = 2/2 = 1
2
x = - 1 - 3 = - 4/2 = - 2
2
R.: x = 1 e x = - 2
(x + 1)² = 3 + x
x² + 2x + 1 = 3 + x
x² + 2x - x + 1 - 3 = 0
x² + x - 2 = 0
a = 1; b = 1; c = - 2
Δ = b² - 4ac
Δ = 1² - 4.1.(-2)
Δ = 1 + 8
Δ = 9
x = - b +/- √Δ = - 1 +/- √9
2a 2.1
x = - 1 + 3 = 2/2 = 1
2
x = - 1 - 3 = - 4/2 = - 2
2
R.: x = 1 e x = - 2
--------------------------------------------------
b)
(t - 1)² + (t + 2)² - 9 = 0
t² - 2t + 1 + t² + 2.2t + 4 - 9 = 0
t² - 2t + 1 + t² + 4t - 5 = 0
t² + t² - 2t + 4t + 1 - 5 = 0
2t² + 2t - 4 = 0 (:2)
t² + t - 2 = 0
a = 1; b = 1; c = - 2
Δ = b² - 4ac
Δ = 1² - 4.1.(-2)
Δ = 1 + 8
Δ = 9
x = - b +/- √Δ = - 1 +/- √9
2a 2.1
x = - 1 + 3 = 2/2 = 1
2
x = - 1 - 3 = - 4/2 = - 2
2
R.: x = 1 e x = - 2
Perguntas interessantes