Odpowiedź :
a) y = -2x²+20x-18
-2x²+20x-18 = 0
a =-2
b =20
c =-18
Δ = b² - 4*a*c
Δ = 20²- 4*(-2)*(-18) = 400 - 144 = 256
√Δ = √256 = 16
x₁ = (-b-√Δ):2a
x₁ = (-20 -16) : 2*(-2) = (-36 ):(- 4 ) = 9
x₁ = 9
x₂ = (-b+√Δ):2a
x₂ = (-20 +16) : 2*(-2) = (-4): (-4) =1
x₂ = 1
y = -2x²+20x-18
y = a(x - x₁ ) ( x- x₂ )
y = -2(x - 9 )(x - 1)
b) y = 3x²-6x+3
3x²-6x+3 = 0
a = 3
b = - 6
c = 3
Δ = b² - 4*a*c
Δ = (-6)² - 4*3*3
Δ = 36 - 36 = o
Δ = 0 więc istnieje 1 podwójny pierwiastek
x₁ = x₂ = -b: 2a
x₁ = x₂ = - (-6) : 2*3
x₁ = x₂ = 6 : 6 =1
x₁ = x₂ = 1
y = 3x²-6x+3
y = 3( x - x₁ )( x - x₁ )
y = 3(x -1)²
-2x²+20x-18 = 0
a =-2
b =20
c =-18
Δ = b² - 4*a*c
Δ = 20²- 4*(-2)*(-18) = 400 - 144 = 256
√Δ = √256 = 16
x₁ = (-b-√Δ):2a
x₁ = (-20 -16) : 2*(-2) = (-36 ):(- 4 ) = 9
x₁ = 9
x₂ = (-b+√Δ):2a
x₂ = (-20 +16) : 2*(-2) = (-4): (-4) =1
x₂ = 1
y = -2x²+20x-18
y = a(x - x₁ ) ( x- x₂ )
y = -2(x - 9 )(x - 1)
b) y = 3x²-6x+3
3x²-6x+3 = 0
a = 3
b = - 6
c = 3
Δ = b² - 4*a*c
Δ = (-6)² - 4*3*3
Δ = 36 - 36 = o
Δ = 0 więc istnieje 1 podwójny pierwiastek
x₁ = x₂ = -b: 2a
x₁ = x₂ = - (-6) : 2*3
x₁ = x₂ = 6 : 6 =1
x₁ = x₂ = 1
y = 3x²-6x+3
y = 3( x - x₁ )( x - x₁ )
y = 3(x -1)²
a)
-2x²+20x-18=-2(x²-10x+9)
Δ=100-36=64
√Δ=8
x₁=(10-8)/2=1
x₂=(10+8)/2=9
dalej mamy -2(x-1)(x-9)
b)
3(x²-2x+1)=3(x-1)²
-2x²+20x-18=-2(x²-10x+9)
Δ=100-36=64
√Δ=8
x₁=(10-8)/2=1
x₂=(10+8)/2=9
dalej mamy -2(x-1)(x-9)
b)
3(x²-2x+1)=3(x-1)²