d) y2 + 2y + 1 = 0
g) x2 - 3x - 130 = 0
j)x^2 + 12x + 32 = 0
me ajudem me enviem com cálculo por favor
Respostas
Resposta:
Explicação passo-a-passo:
D) y² + 2y + 1 = 0
a= 1 b= 2 c= 1
-Calculo-
x=-b ±√b²-4..a.c/2.a
x=-2 ±√2²-4..1.1/2.1
x=-2 ±√4-4/2
x=-2 ±√o/2
x=-2 ±0/2 ------------->x= -2 +0/2 = -2/2 = -1
x= -2 -0/2 = -2/2 = -1
=================================================================
G) x² - 3x - 130 = 0
a= 1 b= -3 c= -130
-Calculo-
x=-b ±√b²-4..a.c/2.a
x=-(-3) ±√(-3)²-4..1.(-130)/2.1
x= -(-3) ±√9-(-520)/2
x= -(-3) ±√9+ 520/2
x= -(-3) ±√529/2
x= -(-3) ± 23/2 ------------->x= -(-3) + 23/2 = 3+ 23/2 = 26/2 = 13
x= -(-3) - 23/2= 3 - 23/2 = -20/2 = -10
=================================================================
J)x² + 12x + 32 = 0
a= 1 b= 12 c= 32
-Calculo-
x=-b ±√b²-4..a.c/2.a
x=-12 ±√12²-4..1.32/2.1
x=-12 ±√144- 128/2
x=-12 ±√16/2
x=-12 ± 4/2 ------------->x= -12+ 4/2 = 8/2 = 4
x= -12 - 4/2= -16/2 = -8