(x - 10)(x - 2) ≤ 160
x² - 12x + 20 - 160 ≤ 0
Решаем кв. ур.
x² - 12x - 140 = 0
а = 1; b = -12; c
= -140
D = b² - 4ac = (-12)² - 4 * 1 * (-140) = 144 + 560 = 704
x1 = - b + √D
= - ( -
12) + √704 = 12
+ 8√11 = 6 + 4√11
2a 2 * 1 2
x2 = - b - √D
= - ( - 12) - √704 = 12 + 8√11 = 6 - 4√11
2a 2 * 1
2
6 - 4√11 6 + 4√11
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Ответ: х ∈ [ 6 - 4√11; 6 + 4√11]