TEM QUE RESOLVER NA FÓRMULA DE BHÁSKARA.
1. Vamos resolver, no conjunto IR, as seguintes equações:
A)x² - 2x= 2x-4
B) x² -2x= x+4
C)x² +10= 9x-10
D)6x² + 3x=1+2x
E)9x² +3x +1= 4x²
F)9x² -1=3x- x²
1. Vamos resolver, no conjunto IR, as seguintes equações:
A)x² - 2x= 2x-4
B) x² -2x= x+4
C)x² +10= 9x-10
D)6x² + 3x=1+2x
E)9x² +3x +1= 4x²
F)9x² -1=3x- x²
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Verified answer
A) x²-2x=2x-4x²-2x-2x+4=0
x²-4x+4=0
Δ=b²-4.a.c
Δ=(-4)²-4.1.4
Δ= 16-16
Δ= 0
x= -b+/-√Δ ÷2.a
x= -(-4)+/-√0 ÷2.1
x= 4+/-0 ÷ 2
x¹= 4+0 ÷ 2= 4/2=2
x²=4-0 ÷2= 4/2=2
b)x²-2x=x+4
x²-2x-x-4=0
x²-3x-4=0
Δ=b²-4.a.c
Δ=(-3)²-4.1.(-4)
Δ= 9+16
Δ= 25
x= -b+/-√Δ÷2.a
x= -(-3)+/-√25÷2.1
x= 3+/-5÷2
x¹= 8÷2=4
x²= -2÷2=-1
c)x²+10=9x-10
x²+10-9x+10=0
x²-9x+20=0
Δ=(-9)-4.1.20
Δ=81-80
Δ=1
x=9+/-1÷2
x¹=10÷2=5
x²=8÷2=4
d)6x²+3x=1+2x
6x²+3x-1-2x=0
6x²+x-1=0
Δ=1²-4.6.(-1)
Δ=1 +24
Δ=25
x= -1+/-5÷12
x¹=4/12 ÷2= 2/6 ÷2= 1/3
x²= -6/12÷2=-3/6÷3=-1/2
e)9x²+3x+1=4x²
9x²+3x+1-4x²=0
5x²+3x+1=0
Δ=3²-4.5.1
Δ=9-20
Δ=-11 ∉R
f)9x²-1=3x-x²
9x²-1-3x+x²=0
10x²-3x-1=0
Δ=(-3)²-4.10.(-1)
Δ=9+40
Δ=49
x= 3+/-7÷20
x¹=10/20÷2=5/10÷5=1/2
x²=-4/20÷2=-2/10÷2=-1/5
x² - 2x - 2x + 4 = 0
x² - 4x + 4 = 0
a = 1 b = - 4 c = + 4
Δ = b² - 4.a.c
Δ = (- 4)² - 4.(1).(+4)
Δ = 16- 16
Δ = 0
x = - b ± √Δ
2.a
x = - (- 4) ± √0
2.1
x = + 4 ± 0
2
x'= 4 + 0 = 4 = 2
2 2
x"= 4 - 0 = 4 = 2
2 2
S[2]
B) x² -2x= x+4
x² - 2x - x - 4 = 0
x² - 3x - 4 = 0
a = 1 b = - 3 c = - 4
Δ = b² - 4.a.c
Δ = (-3)² - 4.(1).(-4)
Δ = 9 + 16
Δ = 25
x = - b ± √Δ
2.a
x = - (- 3) ± √25
2.1
x = + 3 ± 5
2
x'= 3 + 5 = 8 = 4
2 2
x"= 3 - 5 = - 2 = - 1
2 2
S[- 1 , 4]
C)x² +10= 9x-10
x² - 9x + 10 + 10 = 0
x² - 9x + 20 = 0
a = 1 b = - 9 c = + 20
Δ = b² - 4.a.c
Δ = (-9)² - 4.(1).(+20)
Δ = 81 - 80
Δ = 1
x = - b ± √Δ
2.a
x = - (- 9) ± √0
2.1
x = + 9 ± 1
2
x'= 9 + 1 = 10 = 5
2 2
x"= 9 - 1 = 8 = 4
2 2
S[4 , 5]
D)6x² + 3x=1+2x
6x² + 3x - 2x - 1 = 0
6x² + x - 1 = 0
a = 6 b = + 1 c = - 1
Δ = b² - 4.a.c
Δ = (1)² - 4.(6).(-1)
Δ = 1 + 24
Δ = 25
x = - b ± √Δ
2.a
x = - (+1) ± √25
2.6
x = -1 ± 5
12
x'= -1 + 5 = + 4 ÷ 4 = 1
12 12 ÷ 4 3
x"= -1 - 5 = - 6 ÷ 6 = - 1
12 12 ÷ 6 2
S[1/3 , -1/2]
E)9x² +3x +1= 4x²
9x² - 4x² + 3x + 1 = 0
5x² + 3x + 1 = 0
a = 5 b = + 3 c = + 1
Δ = (+3)² - 4.(5).(+1)
Δ = 9 - 20
Δ = - 11
Delta negativo, não existe raiz real.
F)9x² -1=3x- x²
9x² + x² - 3x - 1 = 0
10x² - 3x - 1 = 0
a = 10 b = - 3 c = - 1
Δ = b² - 4.a.c
Δ = (-3)² - 4.(10).(-1)
Δ = 9 + 40
Δ = 49
x = - b ± √Δ
2.a
x = - (- 3) ± √49
2.10
x = + 3 ± 7
20
x'= 3 + 7 = 10 ÷ 10 = 1
20 20 ÷ 10 2
x"= 3 - 7 = - 4 ÷ 4 = - 1
20 20 ÷ 4 5
S[- 1/5 , 1/2]