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a) x[tex]x^{2} - 3x - 18 = 0[/tex]
b) 5x - [tex]x^{2}[/tex] = x + 4
c) (x + 3)(2x - 5) = 14x - 8
d) 0 = -19x + 20 - [tex]x^{2}[/tex] e) 3x + 2[tex]x^{2}[/tex] = 4



Odpowiedź :

a) x² - 3x - 18 = 0

Δ = b² - 4ac

Δ = (-3)² - 4 × 1 × (-18) = 9 + 72 = 81

√Δ = √81 = 9

[tex]x_{1} = \frac{3 + 9}{2*1} = \frac{12}{2} = 6[/tex]

[tex]x_{2} = \frac{3 - 9}{2*1} = \frac{-6}{2} =-3[/tex]

b) 5x - x² = x + 4

- x² + 5x - x - 4 = 0

- x² + 4x - 4 = 0

Δ = 16 - 16 = 0

[tex]x = \frac{-4}{2*(-1)} = 2[/tex]

c) (x + 3)(2x - 5) = 14x - 8

2x² - 5x + 6x -15 = 14x - 8

2x² +x - 15 - 14x + 8 = 0

2x² - 13x - 7 = 0

Δ = 169 + 56 = 225

√Δ = 15

[tex]x_{1} = \frac{13 + 15}{4} = 7\\x_{2} = \frac{13 - 15}{4} = -\frac{1}{2}[/tex]

d) 0 = - 19x + 20 - x²

x² + 19x - 20 = 0

Δ = 361 +80 = 441

√Δ = 21

[tex]x_{1} = \frac{-19 + 21}{2} = 1\\ x_{2} = \frac{-19-21}{2} = -20[/tex]

e) 3x + 2x² = 4

2x² + 3x - 4 = 0

Δ = 9 + 32 = 41

√Δ = √41

[tex]x_{1} = \frac{-3 - \sqrt{41} }{4} \\x_{2} = \frac{-3+\sqrt{41} }{4}[/tex]