Question

resolva os conjuntos IR as seguintes equaçoes do 2° grau.
a)(x +5)(x-5) + 41=8x
b)3t(t+1)=2(5t-1)
c)(x+4)(x-3) -14=(1-x)(x-2)

Answer 1

a) (x +5) (x -5) +41 = 8x
x² -5x +5x -25 +41 -8x = 0
x² -8x +16 = 0
∆= b² -4.a.C
∆= (-8)² -4.1.16
∆= 64 - 64
∆= 0

Bhaskara: 8 ±0/2
n' = 8/2 = 4
n" = 8/2 = 4

b) 3t (t +1) = 2(5t -1)
3t² +3t = 10t -2
3t² +3t -10t +2 = 0
3t² -7t +2 = 0
∆= (-7)² -4.3.2
∆= 49 - 24
∆= 25

Bhaskara: 7 ±5/6
t' = 12/6 = 2
t" = 2/6...simplificar por 2
1/3

c) (x +4) (x -3) -14 = (1 -x)(x -2)
x² -3x +4x -12 -14 = x -2 -x² +2x
x² -3x +4x -12 -14 -x +2 +x² -2x = 0
2x² -2x -24 = 0
∆= b² -4.a.C
∆= (-2)² -4. 2.(-24)
∆= 4 + +192
∆= 196

Bhaskara: 2 ± 14/4
x' = 16/4 = 4
x"= -12/4 = -3

Espero ter ajudado!