3.2.2- Zadanie 3.2.2: Kurs maturalny z matematyki rozszerzonej

Rozwiązanie

$m x^{2} - (3 - m) x + m - 3 = 0$
$(1) m \neq = 0$
$(2) Δ > 0 \to m \in (- 1, 3)$

$(3) (x_{1} - x_{2})^{2} > 12$
$x_{1}^{2} - 2 x_{1} x_{2} + x_{2}^{2} > 12$
$x_{1}^{2} + 2 x_{1} x_{2} - 4 x_{1} x_{2} + x_{2}^{2} > 12$
$(x_{1} + x_{2})^{2} - 4 x_{1} x_{2} > 12$
$(\frac{3 - m}{m})^{2} - 4 \cdot \frac{m - 3}{m} > 12$
$\frac{9 - 6 m + m ^{2}}{m ^{2}} - 4 \cdot \frac{m - 3}{m} > 12 ∣ \cdot m^{2}$
$9 - 6 m + m^{2} - 4 m (m - 3) > 12 m^{2}$
$9 - 6 m + m^{2} - 4 m^{2} + 12 m - 12 m^{2} > 0$
$- 15 m^{2} + 6 m + 9 > 0$
$Δ_{m} = 6^{2} - 4 \cdot (- 15) \cdot 9 = 576$
$Δ_{m} = 24$