Question

3. Решите систему неравенств:
х- 4 > 0,
x2- 2х - 15< 0​

Answer 1

$x - 4 > 0 \\ x > - 4 \\ ( - 4 \: .\: + \infty )$

${x}^{2} - 2x - 15 < 0 \\ d = ({ - 2})^{2} - 4 \times 1 \times ( - 15) = 4 + 60 = 64 \\ x1 = \frac{2 + \sqrt{64} }{2 \times 1} = \frac{2 + 8}{2} = \frac{10}{2} = 5 \\ x 2 = \frac{2 - \sqrt{64} }{2} = \frac{2 - 8}{2} = \frac{ - 6}{2} = - 3 \\ ( - 3 \: \: . \: \: 5)$