Question

6. Розв'язати нерiвнiсть: -x² > 2x-3. X​

Answer 1

$- {x}^{2} \geqslant 2x - 3 \\ - {x}^{2} - 2x + 3 \geqslant 0 \: \: | \times ( - 1) \\ {x}^{2} + 2x - 3 \leqslant0$

${x}^{2} + 2x - 3 = 0$

По теореме Виета:

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

$x_{1} + x_{2} = - 2\\ x_{1} x_{2} = - 3 \\ x_{1} = - 3 \\ x_{2} = 1\\ \\ {ax}^{2} + bx + c = a(x - x_{1})(x - x_{2}) \\ {x}^{2} + 2x - 3 = (x +3 )(x - 1) \\ \\ (x + 3)(x - 1) \leqslant 0 \\ + + + [ - 3] - - - [1] + + + \\ x \: \epsilon\: [ - 3; \: 1]$