Question

prosze o zrobienie tegk mega krotki

Answer 1

Odpowiedź:

1.

y = 1/3(x - 3)(x + 4)

Trójmian kwadratowy jest przedstawiony w postaci iloczynowej

y= a(x - x₁)(x - x₂) ,gdzie x₁ i x₂ są pierwiastkami trójmianu

x₁ = 3 , x₂ = - 4

x₁ * x₂ = - 3 * 4 = - 12

Odp: A

2.

f(x) = 4x(x -2√3) = 4x² - 8√3x

a = 4 , b = - 8√3 , c = 0

Δ = b² - 4ac = (- 8√3)² - 4 * 4 * 0 = 64 * 3 = 192

√Δ = √192 = √(64 * 3) = 8√3

x₁ = ( - b - √Δ)/2a= (8√3 - 8√3)/8= 0/8 = 0

x₂ = (-b + √Δ)/2a = (8√3 + 8√3)/8 = 16√3/8 = 2√3

Odp: D

3.

a)

y = - 4(x + 7)(x - 2)

x₁ = - 7 , x₂ = 2

b)

y = √3/2(x + 3√2)(x - 1 - √3) = √3/2(x + 3√2)[(x + (1 + √3)]

x₁= - 3√2 , x₂ = - (1 + √3)

4.

a)

y = - 3x² - x+ 2

a = - 3 , b = - 1 , c = 2

Δ = b² - 4ac = (- 1)² - 4 * (- 3) * 2 = 1 + 12 * 2 = 1 + 24 = 25

√Δ = √25 = 5

x₁ = ( - b - √Δ)/2a = ( 1 - 5)/(- 6) = - 4/(- 6) = 4/6 = 2/3

x₂ = (- b + √Δ)/2a = (1 + 5)/(- 6) = - 6/6 = - 1

Postać iloczynowa

y = a(x - x₁)(x - x₂) = - 3(x - 2/3)(x + 1)

b)

y = 1/2x² - √5x + 2

a = 1/2 , b = - √5 , c = 2

Δ = b² - 4ac = (- √5)² - 4 * 1/2 * 2 = 5 - 2 * 2 = 5 - 4 = 1

√Δ = √1 = 1

x₁ = ( - b - √Δ)/2a = (√5 - 1)/(2 * 1/2) = (√5 - 1)/1 = √5 - 1

x₂ = (- b + √Δ)/2a = (√5+ 1)/(2 * 1/2) = (√5 + 1)/1 = √5 + 1

Postać iloczynowa

y = a(x - x₁)(x - x₂) = 1/2[x - (√5 - 1)][x - (√5 + 1)] = 1/2(x - √5 + 1)(x - √5 - 1)