Oblicz miejsca zerowe funkcji
[tex]\left. \begin{array} { l } { y = 2 x ^ { 2 } + x - 1 } \\ { y = 3 x ^ { 2 } - 9 x } \\ { y = - x ^ { 2 } + 5 x + 6 } \end{array} \right.[/tex]

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Odpowiedź :

Odpowiedź:

1.

y = 2x² + x - 1

2x² + x - 1 = 0

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

Δ = b² - 4ac = 1² - 4 * 2 * (- 1) = 1 + 8 = 9

√Δ = √9 = 3

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

x₂ = ( - b + √Δ)/2a = (- 1 + 3)/4 = 2/4 = 1/2

x₀ = { - 1 , 1/2 }

2.

y = 3x² - 9x

3x² - 9x = 0

3x(x - 3) = 0

3x = 0 ∨ x - 3 = 0

x = 0 ∨ x = 3

x₀ = { 0 , 3 }

3.

y = - x² + 5x + 6

- x² + 5x + 6 = 0

a = - 1 , b = 5 , c = 6

Δ = b² - 4ac = 5² - 4 * (- 1) * 6 = 25 + 24 = 49

√Δ = √49 = 7

x₁ = ( - b - √Δ)/2a = (- 5 - 7)/(- 2) = - 12/(- 2) = 12/2 = 6

x₂ = ( - b + √Δ)/2a = (- 5 + 7)/(- 2) = 2/(- 2) = - 2/2 = - 1

x₀ = { - 1 , 6 }