Question

Знайдіть множину роз'вязків нерівності: 1) x² – 7x + 12 ≤ 0; 2) x2 – 2x – 24 > 0; 3) –x² – x + 6 ≥ 0; 4) –x2 + 3x + 10 < 0

Answer 1

1)

${x}^{2} - 7x + 12 \leqslant 0$

По теореме Виета

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

${x}^{2} - 7x + 12 = 0 \\ x_{1}+x_{2}=7\\ x_{1}x_{2} =12\\ x_{1} =3\\ x_{2} =4 \\ (x - 3)(x - 4) \leqslant 0 \\ + + + + [3] - - - - [4] + + + + \\ x \: \in \: [3 \: ; \: 4]$

2)

${x}^{2} - 2x - 24 > 0$

По теореме Виета:

${x}^{2} - 2x - 24 = 0 \\ x_{1}+x_{2}=2\\ x_{1}x_{2} = - 24\\ x_{1} =6\\ x_{2} = - 4 \\ (x - 6)(x + 4) > 0 \\ + + + + ( - 4) - - - - (6) + + + + \\ x \: \in \: ( - \infty ; \: - 4)U(6; \: + \infty )$

3)

$- {x}^{2} - x + 6 \geqslant 0 \\ {x}^{2} + x - 6 \leqslant 0$

По теореме Виета:

${x}^{2} + x - 6 = 0 \\ x_{1}+x_{2}= - 1\\ x_{1}x_{2} = - 6\\ x_{1} = - 3\\ x_{2} =2 \\ (x + 3)(x - 2) \leqslant 0 \\ + + + + [ - 3] - - - - [2] + + + + \\ x \:\in \: [ - 3; \: 2]$

4)

$- {x}^{2} + 3x + 10 < 0 \\ {x}^{2} - 3x - 10 > 0$

По теореме Виета:

$x_{1}+x_{2}=3\\ x_{1}x_{2} = - 10\\ x_{1} = - 2\\ x_{2} =5 \\ (x +2 )(x - 5) > 0 \\ + + + + ( - 2) - - - - (5) + + + + \\ x \:\in \: ( - \infty ; \: - 2)U(5; \: + \infty )$