Rozwiąż nierówność x^2+2x-8 lub

zad3 f(x) = x^2 - 3x + 5 f(x) = 4 x^2 - 3x + 5 = 4 x^2 - 3x + 1 = 0 a = 1 \quad b = -3 \quad c = 1 \Delta = b^2 - 4ac = (-3)^2-4 \cdot 1 \cdot 1 = 9 - 4 = 5 x_1 = \frac {-b - \sqrt{\Delta}}{2a} = \frac {3 - \sqrt{5}}{2} x_2 = \frac {-b + \sqrt{\Delta}}{2a} = \frac {3 + \sqrt{5}}{2}

Zad 1. Rozwiąż nierówność x^2 + 2x - 8 < 0 x^2 + 2x - 8 = 0 a = 1 \quad b = 2 \quad c = -8 \Delta = b^2 - 4ac = 2^2-4 \cdot 1 \cdot (-8) = 4 + 32 = 36 x_1 = \frac {-b - \sqrt{\Delta}}{2a} = \frac {-2 - \sqrt{36}}{2} = \frac {-2 - 6}{2} = -4 x_2 = \frac {-b + \sqrt{\Delta}}{2a} = \frac {-2 + \sqrt{36}}{2} = \frac {-2 + 6}{2} = 2 x \in (-4 ; 2) Podaj przykład nierówności: x \in (-\infty;-4> \cup

Zad 2. A: x^2 - 2x - 24 \leqslant 0 x^2 - 2x - 24 = 0 a = 1 \quad b = -2 \quad c = -24 \Delta = b^2 - 4ac = (-2)^2-4 \cdot 1 \cdot (-24) = 4 + 96 = 100 x_1 = \frac {-b - \sqrt{\Delta}}{2a} = \frac {2 - \sqrt{100}}{2} = \frac {2 - 10}{2} = -4 x_2 = \frac {-b + \sqrt{\Delta}}{2a} = \frac {2 + \sqrt{100}}{2} = \frac {2 + 10}{2} = 6 x \in B: |x - a| \leqslant b x \in b = \frac{-4 + 6}{2}=1 a = 6 - b = 6 - 1 = 5 albo a = b - (-4) = 1 + 4 = 5 Czyli: a = 5, \quad b = 1