Maths

2012-09-21 7:47 am
7n^2+9n-500>0,求n的值。

回答 (2)

2012-09-21 8:11 am
✔ 最佳答案
Method 1:
7n² + 9n - 500 > 0
7n² + 9n > 500
7[n² + (9/7)] > 500
7[n² + (9/7) + (9/14)²] > 500 + 7*(9/14)²
7[n + (9/14)]² > 14081/28
n + (9/14) > (√14081)/14 or n + (9/14) <-(√14081)/14
n > (-9 + √14081)/14 or n < (-9 - √14081)/14


Method 2:
7n² + 9n - 500 > 0
7 [n - (-9 + √14081)/14] [n - (-9 - √14081)/14] > 0
n > (-9 + √14081)/14 or n < (-9 - √14081)/14
參考: micatkie
2012-09-22 1:06 am
micatkie個答案有問題 5係乘負數只係加減5洗轉符號
n > (-9 + √14081)/14 or n > (-9 - √14081)/14
之後
n > (-9 + √14081)/14 > (-9 - √14081)/14
所以
n > (-9 + √14081)/14 咁先岩


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