Odpowiedzi

2009-05-05T21:10:19+02:00
1) 2x^2+3x+1=0
Δ = 9 - 4*2*1 = 1
X1 = -3-1 / 4 = -1
x2 = -3+1 / 4 = -1/2

2) (x+2)^2=(2x+3)^2-1
x^2 + 4x + 4 = 4x^2 +12x + 9 - 1
3x^2 +8x +4 = 0
Δ = 64 - 4*3*4 = 64-48 = 16
x1 = -8 -4 / 6 = -2
x2 = -8 +4 / 6 = -4/6 = -2/3

3) (x+4) (x+3) = x+3
x^2 + 3x + 4x + 12 = x +3
x^2 +6x +9 = 0
Δ= 36 - 4*9 = 0
x = -6/2 = -3

4) 4x^2-9x= 0
x(4x - 9) = 0
x1 = 0
x2 = 9/4

5) -25x^2+1= 0
25x^2 = 1
x^2 = 1/25
x1 = 1/5
x2 = -1/5
2009-05-05T21:10:36+02:00
1)

2x² + 3x + 1 = 0
Δ = 9 - 8 = 1 √Δ = 1
x₁ = -3 - 1 ÷ 4 = -1
x₂ = -3 + 1 ÷ 4 = -½

2)

(x + 2)² = (2x + 3)²-1 => x² +4x +4 = 4x² + 12x + 9 -1 =>
4x² - x² + 12x - 4x + 8 - 4 = 0 => 3x² + 8x +4 = 0
Δ = 64 - 48 = 16 √Δ = 4
x₁ = -8 - 4 ÷ 6 = -2
x₂ = -8 + 4 ÷ 6 = -⅔

3)

(x + 4) × (x + 3) = x + 3 => (x + 4) × (x + 3) - (x +3) = 0 =>
(x + 3)(x + 4 -1) = 0 => (x + 3)(x + 3) = 0 => x = -3

4)

4x² - 9x = 0 => x(4x - 9)=0 => x = 0 lub 4x - 9 = 0 =>
x = 0 lub x = 2¼

5)

-25x² + 1= 0 => 25x² - 1 = 0 => (5x + 1)(5x - 1) = 0 =>
x = -⅕ lub x = ⅕
2009-05-05T21:14:31+02:00
1)
2x^2+3x+1=0
Δ=3^2-4*1*2
Δ=9-8
Δ=1

x1=-3-1/4=>x1=-1
x2=-3+1/4=>x2=-0,5

2)
(x+2)^2=(2x+3)^2-1
x^2+4x+4=4x^2+12x+9-1

3x^2+8x+4=0
Δ=64-4*4*3
Δ=16
√Δ=4

x1=-8-4/6=>2
x2=-8+4/6=>-(2/3)


3)
(x+4) (x+3) = x+3
x^2+3x+4x+12=x+3

x^2+6x+9=0
Δ=6^2-4*9
Δ=36-36
Δ=0
x0=-6/2
x0=-3


4)
4x^2-9x= 0
Δ=81
√Δ=9
x0=9+9/8
x0=2,25

5)

-25x^2+1= 0
Δ=4*25
Δ=100
x1=-100/-50=>2
x2=100/-50=>-2


  • Użytkownik Zadane
2009-05-05T21:19:07+02:00
2x^2+3x+1=0
Δ=9-4*2*1=9-8=1
√Δ=1
X1=-3-1 PRZEZ 4= -4/4=-1
X2=-3+1 PRZEZ 4=-2/4=-1/2
_________________________
(x+2)^2=(2x+3)^2-1
X²+4X+4=4X²+12X+9-1
X²+4X+4-4X²-12X-9+1=0
-3X²-8X²-4=0
Δ=64-4*(-3)*(-4)=64*12*(-4)=64*(-48)=-3072
Δ<0 BRAK PIERWIASTKÓW
__________________________
(x+4) (x+3) = x+3
X²+3X+4X+12-X-3=0
X²+6X+9=0
Δ=36-4*1*9=36-36=0
X0=-6 PREZ 2= -3, CZYLI (X+3)²
_________________________
4x^2-9x= 0
Δ=-4*4*(-9)=-16*(-9)=144
√Δ=12
X1=-12/4=-3
X2=12/4=3
_____________________________
-25x^2+1= 0
Δ=-4*(-25)*1=100
√Δ=10
X1=-10/2*(-25)=-10/-50=1/5
X2=10/-50=-1/5