|2x-1| ≥ 7?
回答 (11)
2017-09-19 9:19 pm
✔ 最佳答案
1.|2x-1| ≥ 7 ----------(1)
2x-1 = 7
2x=8
x= 4
2x-1 = -7
2x= -6
x= -3
x=-3; x=4
plot the points on the real line.
------ ------ ------- --------
-∞....-3.....4.....∞
Consider the intervals (-∞,-3),(-3,4),(4,∞)
Choose 'any' one point from each interval and test inequality (1)
(-∞, -3) : choose x=-4
|2x-1| ≥ 7
9 ≥ 7 (true)
(-3,4) : choose x=0
|2x-1| ≥ 7
1 ≥ 7 (false)
(4,∞): choose x=5
|2x-1| ≥ 7
9 ≥ 7 (true)
(-∞,-3] U [4, ∞)
2017-09-19 9:47 pm
|2x - 1| ≥ 7
|2x - 1|² ≥ 7²
4x² - 4x + 1 ≥ 49
4x² - 4x - 48 ≥ 0
x² - x - 12 ≥ 0
(x + 3)(x - 4) ≥ 0
x ≤ -3 or x ≥ 4
|2x - 1|² ≥ 7²
4x² - 4x + 1 ≥ 49
4x² - 4x - 48 ≥ 0
x² - x - 12 ≥ 0
(x + 3)(x - 4) ≥ 0
x ≤ -3 or x ≥ 4
2017-09-19 9:35 pm
|2x-1| ≥ 7 ---> 2x - 1 ≤ -7, and 2x - 1 ≥ 7.
A) 2x - 1 ≤ -7 ---> 2x ≤ -7 + 1 ---> x ≤ (-7 + 1)/2 ---> x ≤ -6/2 ---> x ≤ -3.
B) 2x - 1 ≥ 7 ---> 2x ≥ 7 + 1 ---> x ≥ (7 + 1)/2 ---> x ≥ 8/2 ---> x ≥ 4.
Answer: x ≤ -3; x ≥ 4.
A) 2x - 1 ≤ -7 ---> 2x ≤ -7 + 1 ---> x ≤ (-7 + 1)/2 ---> x ≤ -6/2 ---> x ≤ -3.
B) 2x - 1 ≥ 7 ---> 2x ≥ 7 + 1 ---> x ≥ (7 + 1)/2 ---> x ≥ 8/2 ---> x ≥ 4.
Answer: x ≤ -3; x ≥ 4.
2017-09-19 9:47 pm
|a|= √(a²) so |2x-1| ≥ 7
→ √((2x-1)²) ≥ 7
→ (2x-1)² ≥ 49
→ 4x²-4x+1 ≥ 49
→ 4(x²-x-12) ≥ 0
→ 4(x+3)(x-4) ≥ 0
→ x ≤ -3 or x ≥ 4
Graph comparing |2x-1| ≥ 7 and 4(x+3)(x-4) ≥ 0: https://www.desmos.com/calculator/glybc4lypc
→ √((2x-1)²) ≥ 7
→ (2x-1)² ≥ 49
→ 4x²-4x+1 ≥ 49
→ 4(x²-x-12) ≥ 0
→ 4(x+3)(x-4) ≥ 0
→ x ≤ -3 or x ≥ 4
Graph comparing |2x-1| ≥ 7 and 4(x+3)(x-4) ≥ 0: https://www.desmos.com/calculator/glybc4lypc
2017-09-19 10:48 pm
|2x - 1| >= 7
2x - 1 >= 7 or 2x - 1 <= -7
2x >= 8 or 2x <= -6
x >= 4 or x <= -3 (interval notation: [-oo, -3] U [4, oo]).
2x - 1 >= 7 or 2x - 1 <= -7
2x >= 8 or 2x <= -6
x >= 4 or x <= -3 (interval notation: [-oo, -3] U [4, oo]).
2017-09-19 9:18 pm
2x - 1 ≥ 7
2x ≥ 8
x ≥ 4
2x - 1 ≦ - 7
2x ≦ - 6
x ≦ - 3
Solution ( - inf, - 3] U [ 4, + inf)
2x ≥ 8
x ≥ 4
2x - 1 ≦ - 7
2x ≦ - 6
x ≦ - 3
Solution ( - inf, - 3] U [ 4, + inf)
2017-09-19 9:14 pm
Are you supposed to graph it?
2017-09-19 9:11 pm
|2x - 1| ≥ 7
An absolute value is always ≥ 0, so your inequality is always true whatever the value of x.
→ Solution = IR
An absolute value is always ≥ 0, so your inequality is always true whatever the value of x.
→ Solution = IR
2017-09-19 9:06 pm
4 <= x <= -3
2017-09-19 11:18 pm
2x - 1 ≤ - 7___or ____ 7 ≤ 2x - 1
2x ≤ -6_____ or_____2x ≥ - 8
x ≤ - 3_____or_______x ≥ - 4
solution set is { - 4 ≤ x ≤ - 3 }
2x ≤ -6_____ or_____2x ≥ - 8
x ≤ - 3_____or_______x ≥ - 4
solution set is { - 4 ≤ x ≤ - 3 }
2017-09-19 9:16 pm
2x-1 ≥ 7 or 2x-1 ≤ -7
2x ≥ 8 or 2x ≤ -6
x ≥ 4 or x ≤ -3
(-∞,-3) U (4,∞)
2x ≥ 8 or 2x ≤ -6
x ≥ 4 or x ≤ -3
(-∞,-3) U (4,∞)
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