Question

Количество целых решений неравенства:
х^2+14х+49<5 |х+7|

Answer 1

${x}^{2} + 14x + 49 < 5 times |x + 7| \ 1)x + 7 geqslant 0 : : : x geqslant -7 \ {x}^{2} + 14x + 49 < 5 times (x + 7) \ {x}^{2} + 14x + 49 < 5x + 35 \ {x}^{2} + 9x + 14 < 0 \ {x }^{2} + 9x + 14 = 0 \ d = 81 - 4 times 14 = 81 - 54 = 25 = {5}^{2} \ x = frac{ - 9±5}{2} = - 2 : : - 7 \ + + + + ( - 7) - - - ( - 2) + + \ xin( - 7; - 2) \ 2)x + 7 < 0 : : : x < - 7 \ {x }^{2} + 14x + 49 < 5 times - (x + 7) \ {x}^{2} + 14x + 49 < - 5x - 35 \ {x}^{2} + 19x + 84 < 0 \ {x}^{2} + 19x + 84 = 0 \ d : = 361 - 4 times 84 = 361 - 336 = 25 = {5}^{2} \ x = frac{ - 19 ±5}{2} = - 12; - 7 \ + + + + ( - 12) - - - ( - 7) + + + \ xin( - 12;- 7)$
Ответ :
$xin( - 12; - 7)cup ( - 7; - 2)$
Количество целых решений : 8