Question

wyznacz miejsca zerowe f kwadratowej y+2xkwadrat-4x+2 y=-xkwadrat+3x+4

Answer 1

y = 2x² - 4x  +2

Δ=b²-4ac = 16 - 16=0

x=-b/2a = 4/4 = 1

 

lub inaczej:

2x² -4x + 2=0

2(x²-2x+1) =0

 2(x-1)²=0  /:2

   (x-1)²=0

    x = 1

odp:

x=1

 

y = -x² + 3x + 4

Δ=9 + 16 = 25

√Δ=5

x1=(-b-√Δ)/2a = (-3-5)/-2 = -8/-2 = 4

x2=(-b+√Δ)/2a = (-3+5)/-2 = 2/-2 = -1

odp:

x = 4   ∨   x=-1

 

Answer 2

a)

y = 2x²-4x+2

2x²-4x+2 = 0  /:2

x²-2x+1 = 0

a = 1,  b = -2,  c = 1

Δ = b²-4ac = 4-4 = 0

x = -b/2a = -(-2)/2 = 2/2 = 1

Odp. MZ: 1

 

b)

y = -x²+3x+4

-x²+3x+4 = 0

a = -1,  b = 3,  c = 4

Δ = b²-4ac = 9+16 = 25

√Δ = √25 = 5

x1 = (-b-√Δ)/2a = (-3-5)/(-2) = 4

x2 = (-b+√Δ)/2a = (-3+5)/(-2) = -1

MZ: 4; -1