Question

x-6)≤(x+1)(x-3)+4 пж​

Answer 1

Ответ: ↓

$x - 6 \leqslant (x + 1)(x - 3) + 4 \\ x - 6 \leqslant {x}^{2} - 3x + x - 3 + 4 \\ x - 6 \leqslant {x}^{2} - 2x + 1 \\ {x}^{2} + 3x - 7 \geqslant 0$

$D = {3}^{2} - 4 \times 1 \times ( - 7) =9 + 28 = 37$

$x1x2 = \frac{ - 3± \sqrt{37} }{2 \times 1} = \frac{ - 3 - \sqrt{37} }{2} \: ; \: \frac{ - 3 + \sqrt{37} }{2}$