Question

x(x-2)>3(x-2)

(1-3x)²+3x-4x²<9

i tu w takim nawiasie razem(czyli w takiej klamerce) uklad rownan:

{y=x²+x+1

y=-6x+9

Answer 1

x(x - 2) > 3(x - 2)

x² - 2x > 3x - 6

x² - 2x - 5x + 6 > 0

x² - 7x + 6 > 0

Δ = b² - 4ac = (-7)² - 4 × 6 × 1 = 49 - 24 = 25

√Δ = √25 = 5

x₁ = (-b - √Δ) ÷ 2a = (7 - 5) ÷ 2 = 2 ÷ 2 = 1

x₂ = (-b + √Δ) ÷ 2a = (7 + 5) ÷ 2 = 12 ÷ 2 = 6

a > 0

x² - 7x + 6 > 0 <=> x ∈ (-∞ ; 1) U (6 ; ∞)

 

(1 - 3x)² + 3x - 4x² < 9

1 - 6x + 9x² + 3x - 4x² - 9 > 0

5x² - 3x - 8 > 0

Δ = b² - 4ac = (-3)² - 4 × 5 × (-8) = 9 + 160 = 169

√Δ = √169 = 13

x₁ = (-b - √Δ) ÷ 2a = (3 - 13) ÷ 2 × 5 = -10 ÷ 10 = -1

x₂ = (-b + √Δ) ÷ 2a = (3 + 13) ÷ 2 × 5 = 16 ÷ 10 = 1⅗

a > 0

(1 - 3x)² + 3x - 4x² < 9 <=> x ∈ (-∞ ; -1) U (1⅗ ; ∞)

 

y = x² + x + 1

y = -6x + 9

 

x² + x + 1 = -6x + 9

x² + x + 1 + 6x - 9 = 0

x² + 7x - 8 = 0

Δ = b² - 4ac = 7² - 4 × (-8) = 49 + 32 = 81

Δ = √81 = 9

x₁ = (-b - √Δ) ÷ 2a = (-7 - 9) ÷ 2 × 1 = -16 ÷ 2 = -8

x₂ = (-b + √Δ) ÷ 2a = (-7 + 9) ÷ 2 × 1 = 2 ÷ 2 = 1

Jeśli chodziło znalezienie punktów wspólnych dla tych funkcji to tyle. Jeśli to układ na dwie niewiadome:

y = -6x + 9

y = -6 × (-8) + 9          ∨        y = -6 × 1 + 9

y = 48 + 9                   ∨        y = -6 + 9

y = 57                        ∨         y = 3