v- Zadanie niewykorzystane2: Kurs maturalny z matematyki rozszerzonej

Rozwiązanie

${x = \frac{16 + 4 k + k ^{2}}{4 + k} y = \frac{- 16 k}{4 + k}$ $∣ x - y ∣ \geq 4$

$\frac{16 + 4 k + k ^{2}}{4 + k} - \frac{- 16 k}{4 + k} \geq 4$
$\frac{16 + 4 k + k ^{2} + 16 k}{4 + k} \geq 4$
$\frac{16 + k ^{2} + 20 k}{4 + k} \geq 4$
$\frac{16 + 4 k + k ^{2} + 16 k}{4 + k} \geq 4 \lor \frac{16 + 4 k + k ^{2} + 16 k}{4 + k} \leq - 4$
$\frac{16 + 4 k + k ^{2} + 16 k}{4 + k} - 4 \geq 0 \frac{16 + 4 k + k ^{2} + 16 k}{4 + k} + 4 \leq 0$
$\frac{16 + 4 k + k ^{2} + 16 k}{4 + k} - \frac{16 + 4 k}{4 + k} \geq 0 \frac{16 + 4 k + k ^{2} + 16 k}{4 + k} + \frac{16 + 4 k}{4 + k} \leq 0$
$\frac{16 + 4 k + k ^{2} + 16 k - 16 - 4 k}{4 + k} \geq 0 \frac{16 + 4 k + k ^{2} + 16 k + 16 + 4 k}{4 + k} \leq 0$
$\frac{k ^{2} + 16 k}{4 + k} \geq 0 \frac{32 + 24 k + k ^{2}}{4 + k} \leq 0$
$(k^{2} + 16 k) (4 + k) \geq 0 (32 + 24 k + k^{2}) (4 + k) \leq 0$
$k (k + 16) (4 + k) \geq 0 (k^{2} + 24 k + 32) (4 + k) \leq 0$
$k (k + 16) (4 + k) = 0$ $(k^{2} + 24 k + 32) (4 + k) = 0$
$k = 0 \lor k = - 16 \lor k = - 4$ $Δ = 2 4^{2} - 4 \cdot 32 = 448, Δ = 87$