solve the following inequalities by the algebraic method.
1. 4x(x+2)<3(x+17)
math question 20point
2011-03-13 3:12 am
回答 (2)
2011-03-13 3:53 am
✔ 最佳答案
============================================================4x(x + 2) < 3(x + 17)
4x^2 + 8x < 3x + 51
4x^2 + 5x - 51 < 0
(4x + 17)(x - 3) < 0
-17/4 < x < 3
2011-03-12 19:54:54 補充：
(4x + 17)(x - 3) < 0
Case 1:
4x + 17 > 0 => x > -17/4
and x - 3 < 0 => x < 3
So we have -17/4 < x < 3
Case 2:
4x + 17 < 0 => x < - 17/4
and x - 3 > 0 => x > 3
So this case is rejectable.
參考: Knowledge is power.
2011-03-13 3:52 am
4x(x + 2) < 3(x + 17)
4x^2 + 8x < 3x + 51
4x^2 + 5x – 51 < 0
(4x + 17)(x – 3) < 0
Since the product of 2 terms is negative, one is +ve and one is –ve
2011-03-12 19:52:23 補充：
(4x + 17 > 0 and x – 3 < 0) or (4x +17 < 0 and x – 3 >0)
-17/4 < x < 3 or (x < -17/4 and x > 3) (rejected)
-17/4 < x < 3
4x^2 + 8x < 3x + 51
4x^2 + 5x – 51 < 0
(4x + 17)(x – 3) < 0
Since the product of 2 terms is negative, one is +ve and one is –ve
2011-03-12 19:52:23 補充：
(4x + 17 > 0 and x – 3 < 0) or (4x +17 < 0 and x – 3 >0)
-17/4 < x < 3 or (x < -17/4 and x > 3) (rejected)
-17/4 < x < 3
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