Question

Потрібен розвязок даю 100 балів
Алгебра

Answer 1

По теореме Виета:

${x}^{2} + bx + c = 0\\ x_{1} + x_{2} = - b\\ x_{1} x_{2} = c$

${x}^{2} - 3x - 4 = 0 \\ x_{1} + x_{2} =3 \\ x_{1} x_{2} = - 4 \\ x_{1} = - 1 \\ x_{2} = 4\\ \\ {ax}^{2} + bx + c = a(x - x_{1})(x - x_{2}) \\ {x}^{2} - 3x - 4 = (x + 1)(x - 4)$

1) Г

$x {}^{2} - 3x - 4 > 0 \\ (x + 1)(x - 4) > 0 \\ + + + ( - 1) - - - (4 )+ + + \\ x \: \epsilon \: ( - \propto; \: - 1)U(4; \: + \propto)$

2) А

${x}^{2} - 3x - 4 < 0 \\ (x + 1)(x - 4) < 0 \\ + + + ( - 1) - - - (4) + + + \\ x \: \epsilon \: ( - 1; \: 4)$

3) Д

${x}^{2} - 3x - 4 \geqslant 0 \\ (x + 1)(x - 4) \geqslant 0 \\ + + + [ - 1] - - - [4] + + + \\ x \: \epsilon \: ( - \propto; \: - 1]U[4; \: + \propto)$

4) В

${x}^{2} - 3x - 4 \leqslant 0 \\ (x + 1)(x - 4) \leqslant 0 \\ + + + [ - 1] - - - [4] + + + \\ x \: \epsilon \: [ - 1; \: 4]$